89:"UserProgressProvider" 8a:I[651451,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$89"] 8b:I[196793,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"Gingerbread"] 8d:"TopicRecentlyViewed" 8e:I[716297,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$8d"] 8f:"RecordResourceResume" 90:I[647935,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$8f"] 91:"TopicQuestionBank" 92:I[426249,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$91"] 93:T476,- Calculates the moles of iodine, $I_2$, reacted in the titration: $n(\text{I}_2) = 1.50 \times 10^{-3}\text{ mol dm}^{-3} \times \frac{18.40}{1000}\text{ dm}^3 = 2.76 \times 10^{-5}\text{ mol}$ M1 - Deduces that the moles of sulfur dioxide, $\text{SO}_2$, in the $25.0\text{ cm}^3$ titrated aliquot is $2.76 \times 10^{-5}\text{ mol}$ (due to $1:1$ reacting ratio) M1 - Scales to find the total moles of sulfur dioxide in the $250.0\text{ cm}^3$ volumetric flask: $n(\text{SO}_2)_{\text{total}} = 2.76 \times 10^{-5}\text{ mol} \times \frac{250.0}{25.0} = 2.76 \times 10^{-4}\text{ mol}$ M1 - Calculates the molar mass of sulfur dioxide, $\text{SO}_2$: $M(\text{SO}_2) = 32.1 + (2 \times 16.0) = 64.1\text{ g mol}^{-1}$ (accepts $64.0\text{ g mol}^{-1}$) M1 - Calculates the mass of $\text{SO}_2$ in the serving in grams: $\text{mass} = 2.76 \times 10^{-4}\text{ mol} \times 64.1\text{ g mol}^{-1} = 0.01769\text{ g}$ (or $0.01766\text{ g}$ if $M_r = 64.0$) M1 - Converts the mass to milligrams and rounds to 3 significant figures: $17.7\text{ mg}$ A194:T497,To solve this problem using **Dimensional Analysis**, we will construct a single, continuous multiplication chain of conversion factors. This chain will start with our measured input value (the volume of iodine solution used) and end with the target unit: **milligrams of sulfur dioxide** ($\text{mg } \text{SO}_2$) in the original $200\text{ cm}^3$ serving. Let's first list our starting values and conversion factors: * **Measured volume of $\text{I}_2$:** $18.40\text{ cm}^3 \text{ I}_2 \text{ solution}$ * **Concentration of $\text{I}_2$:** $1.50 \times 10^{-3}\text{ mol dm}^{-3}$ (which means $\frac{1.50 \times 10^{-3}\text{ mol } \text{I}_2}{1000\text{ cm}^3\text{ solution}}$) * **Stoichiometric ratio:** $\frac{1\text{ mol } \text{SO}_2}{1\text{ mol } \text{I}_2}$ (from the balanced equation) * **Dilution factor:** $\frac{250.0\text{ cm}^3\text{ total solution}}{25.0\text{ cm}^3\text{ titrated sample}}$ * **Molar mass of $\text{SO}_2$:** $32.1 + 2 \times 16.0 = 64.1\text{ g mol}^{-1}$ (using standard periodic table values, $\frac{64.1\text{ g } \text{SO}_2}{1\text{ mol } \text{SO}_2}$) * **Unit conversion:** $\frac{1000\text{ mg}}{1\text{ g}}$95:T51e,Now we arrange these terms so that all intermediate units cancel out, leaving us with only the desired unit: $\text{mg } \text{SO}_2$. $$\text{Mass of } \text{SO}_2 \text{ (mg)} = 18.40\text{ cm}^3 \text{ I}_2 \text{ soln} \times \left( \frac{1.50 \times 10^{-3} \text{ mol } \text{I}_2}{1000\text{ cm}^3 \text{ I}_2 \text{ soln}} \right) \times \left( \frac{1 \text{ mol } \text{SO}_2}{1 \text{ mol } \text{I}_2} \right) \times \left( \frac{250.0\text{ cm}^3 \text{ total}}{25.0\text{ cm}^3 \text{ sample}} \right) \times \left( \frac{64.1 \text{ g } \text{SO}_2}{1 \text{ mol } \text{SO}_2} \right) \times \left( \frac{1000 \text{ mg}}{1 \text{ g}} \right)$$ Notice how the units cancel step-by-step: * $\text{cm}^3 \text{ I}_2 \text{ soln}$ cancels with $\text{cm}^3 \text{ I}_2 \text{ soln}$ in the concentration denominator. * $\text{mol } \text{I}_2$ cancels with $\text{mol } \text{I}_2$ in the mole ratio denominator. * $\text{mol } \text{SO}_2$ cancels with $\text{mol } \text{SO}_2$ in the molar mass denominator. * The dilution ratio accounts for scaling from the $25.0\text{ cm}^3$ aliquot back to the total $250.0\text{ cm}^3$ flask volume. * $\text{g}$ cancels with $\text{g}$ in the milligram conversion denominator. We are left with the final units of **$\text{mg } \text{SO}_2$**.96:T497,To solve this problem using **Dimensional Analysis**, we will construct a single, continuous multiplication chain of conversion factors. This chain will start with our measured input value (the volume of iodine solution used) and end with the target unit: **milligrams of sulfur dioxide** ($\text{mg } \text{SO}_2$) in the original $200\text{ cm}^3$ serving. Let's first list our starting values and conversion factors: * **Measured volume of $\text{I}_2$:** $18.40\text{ cm}^3 \text{ I}_2 \text{ solution}$ * **Concentration of $\text{I}_2$:** $1.50 \times 10^{-3}\text{ mol dm}^{-3}$ (which means $\frac{1.50 \times 10^{-3}\text{ mol } \text{I}_2}{1000\text{ cm}^3\text{ solution}}$) * **Stoichiometric ratio:** $\frac{1\text{ mol } \text{SO}_2}{1\text{ mol } \text{I}_2}$ (from the balanced equation) * **Dilution factor:** $\frac{250.0\text{ cm}^3\text{ total solution}}{25.0\text{ cm}^3\text{ titrated sample}}$ * **Molar mass of $\text{SO}_2$:** $32.1 + 2 \times 16.0 = 64.1\text{ g mol}^{-1}$ (using standard periodic table values, $\frac{64.1\text{ g } \text{SO}_2}{1\text{ mol } \text{SO}_2}$) * **Unit conversion:** $\frac{1000\text{ mg}}{1\text{ g}}$97:T51e,Now we arrange these terms so that all intermediate units cancel out, leaving us with only the desired unit: $\text{mg } \text{SO}_2$. $$\text{Mass of } \text{SO}_2 \text{ (mg)} = 18.40\text{ cm}^3 \text{ I}_2 \text{ soln} \times \left( \frac{1.50 \times 10^{-3} \text{ mol } \text{I}_2}{1000\text{ cm}^3 \text{ I}_2 \text{ soln}} \right) \times \left( \frac{1 \text{ mol } \text{SO}_2}{1 \text{ mol } \text{I}_2} \right) \times \left( \frac{250.0\text{ cm}^3 \text{ total}}{25.0\text{ cm}^3 \text{ sample}} \right) \times \left( \frac{64.1 \text{ g } \text{SO}_2}{1 \text{ mol } \text{SO}_2} \right) \times \left( \frac{1000 \text{ mg}}{1 \text{ g}} \right)$$ Notice how the units cancel step-by-step: * $\text{cm}^3 \text{ I}_2 \text{ soln}$ cancels with $\text{cm}^3 \text{ I}_2 \text{ soln}$ in the concentration denominator. * $\text{mol } \text{I}_2$ cancels with $\text{mol } \text{I}_2$ in the mole ratio denominator. * $\text{mol } \text{SO}_2$ cancels with $\text{mol } \text{SO}_2$ in the molar mass denominator. * The dilution ratio accounts for scaling from the $25.0\text{ cm}^3$ aliquot back to the total $250.0\text{ cm}^3$ flask volume. * $\text{g}$ cancels with $\text{g}$ in the milligram conversion denominator. We are left with the final units of **$\text{mg } \text{SO}_2$**.72:["$","$L8a",null,{"bundle":{"topicIds":["202b3280-0ab3-5a7f-9232-dc34b579e594","d88fc8b9-7df9-5b0e-9ff0-804962981bff","333f36fe-88c6-5c1a-bbc5-5533bc34825e","673ce67d-a4bd-55fe-af5e-da5ec80acfa3","f555f6a4-d798-5459-927e-e475517e3160","b9ae8f74-1d53-53d6-ba98-a3aee324f0ed"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L8b",null,{"feature":"question-bank"}],"$L8c",["$","$L8e",null,{"subjectCode":"a_level_chemistry__ocr_a","subjectName":"Chemistry","topicSlug":"atoms-and-reactions","topicTitle":"Atoms and reactions","sectionTitle":null,"activity":"worksheet","resourceTitle":"Atoms and reactions Questions","resourceHref":"/a-level/chemistry/ocr/a/atoms-and-reactions/questions/q1","saveTrail":false}],["$","$L90",null,{"href":"/a-level/chemistry/ocr/a/atoms-and-reactions/questions/q1","label":"Atoms and reactions","subjectCode":"a_level_chemistry__ocr_a","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 79 – Atoms and reactions - Chemistry Questions\",\"description\":\"Practise Atoms and reactions A Level Chemistry exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"Atoms and reactions\"},\"url\":\"https://mathsgenie.co.uk/a-level/chemistry/ocr/a/atoms-and-reactions/questions/q79\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L92",null,{"backHref":"/a-level/chemistry/ocr/a/questions","basePath":"/a-level/chemistry/ocr/a/atoms-and-reactions/questions","formulaUrl":"https://www.ocr.org.uk/Images/363792-unit-h032-and-h432-data-sheet.pdf","initialIndex":78,"initialQuestionWindow":[{"id":"67fb5c73-a004-47d7-b8d2-2d9ada07bc57","revisionId":"0e27b518-c7b3-4762-a348-281e532af554","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"d86c281f-7463-466f-9620-23d748d224bd","label":null,"content":"Vanadium(V) oxide, \$\\text{V}_2\\text{O}_5\$, is reduced to vanadium metal by heating with calcium. What is the minimum mass of \$\\text{Ca}\$ required to reduce \$45.45\\text{ g}\$ of vanadium(V) oxide?\n\n(Use the relative atomic masses: \$\\text{V} = 50.9\$, \$\\text{Ca} = 40.1\$, \$\\text{O} = 16.0\$)","metadata":null,"markscheme":"- Writes or uses the balanced equation: $\\text{V}_2\\text{O}_5 + 5\\text{Ca} \\rightarrow 2\\text{V} + 5\\text{CaO}$ (or otherwise identifies the $1:5$ molar ratio between $\\text{V}_2\\text{O}_5$ and $\\text{Ca}$) M1\n- Calculates the molar mass of $\\text{V}_2\\text{O}_5$: $M_r = (2 \\times 50.9) + (5 \\times 16.0) = 181.8\\text{ g mol}^{-1}$ M1\n- Calculates the amount in moles of $\\text{V}_2\\text{O}_5$: $n = \\frac{45.45}{181.8} = 0.250\\text{ mol}$ A1\n- Determines the amount in moles of $\\text{Ca}$ required: $n(\\text{Ca}) = 5 \\times 0.250 = 1.25\\text{ mol}$ M1\n- Calculates the minimum mass of $\\text{Ca}$ needed: $\\text{mass} = 1.25 \\times 40.1 = 50.1\\text{ g}$ (accept $50.125\\text{ g}$) A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"stepwise_mole_method","label":"Stepwise Mole Calculation","steps":[{"type":"text","order":0,"title":"Write the balanced equation","content":"To find the relationship between vanadium(V) oxide ($\\text{V}_2\\text{O}_5$) and calcium ($\\text{Ca}$), we must first write a balanced chemical equation for the reduction reaction:\n\n$$\\text{V}_2\\text{O}_5 + 5\\text{Ca} \\rightarrow 2\\text{V} + 5\\text{CaO}$$\n\nThis shows us that the stoichiometric ratio between $\\text{V}_2\\text{O}_5$ and $\\text{Ca}$ is **$1:5$**.","highlights":[{"text":"Vanadium(V) oxide, \$\\text{V}_2\\text{O}_5\$, is reduced to vanadium metal by heating with calcium.","location":"specification"}]},{"type":"text","order":1,"title":"Calculate molar mass of V2O5","content":"Next, we calculate the molar mass ($M_r$) of vanadium(V) oxide, $\\text{V}_2\\text{O}_5$, using the given relative atomic masses ($\\text{V} = 50.9$ and $\\text{O} = 16.0$):\n\n$$\\begin{aligned} M_r(\\text{V}_2\\text{O}_5) &= (2 \\times 50.9) + (5 \\times 16.0) \\\\ &= 101.8 + 80.0 \\\\ &= 181.8\\text{ g mol}^{-1} \\end{aligned}$$","highlights":[{"text":"\\text{V} = 50.9","location":"specification"},{"text":"\\text{O} = 16.0","location":"specification"}]},{"type":"text","order":2,"title":"Calculate moles of V2O5","content":"Now, we calculate the chemical amount in moles of $\\text{V}_2\\text{O}_5$ present in $45.45\\text{ g}$ of the substance using the formula:\n\n$$\\text{moles} = \\frac{\\text{mass}}{\\text{molar mass}}$$\n\n$$\\begin{aligned} n(\\text{V}_2\\text{O}_5) &= \\frac{45.45\\text{ g}}{181.8\\text{ g mol}^{-1}} \\\\ &= 0.250\\text{ mol} \\end{aligned}$$","calculator":[{"gdc":{"expression":"45.45 / 181.8","instructions":"Enter 45.45 / 181.8 in the calculation screen and press EXE."},"ti84":{"expression":"45.45 / 181.8","instructions":"Enter 45.45 divided by 181.8 on the home screen and press ENTER."},"desmos":{"expression":"45.45 / 181.8","instructions":"Type 45.45 / 181.8 in the input line."},"description":"Calculate the moles of vanadium(V) oxide by dividing the mass by the molar mass."}],"highlights":[{"text":"45.45\\text{ g}","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{molar ratio of } \\text{V}_2\\text{O}_5 : \\text{Ca} = \\htmlData{anchor=ratio-v}{\\color{@sky}{1}} : \\htmlData{anchor=ratio-ca}{\\color{@amber}{5}}$"},{"latex":"$$n(\\text{Ca}) = \\htmlData{anchor=factor-ca}{\\color{@amber}{5}} \\times \\htmlData{anchor=moles-v}{\\color{@sky}{0.250}}\\text{ mol} = \\htmlData{anchor=moles-ca}{\\color{@purple}{1.25}}\\text{ mol}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"ratio-v"},{"color":"amber","style":"text-color","anchor":"ratio-ca"},{"color":"sky","style":"text-color","anchor":"moles-v"},{"color":"purple","style":"text-color","anchor":"moles-ca"}],"connections":[{"to":"factor-ca","from":"ratio-ca","color":"amber","label":"multiply factor","style":"solid"},{"to":"moles-ca","from":"moles-v","color":"purple","label":null,"style":"dashed"}]},"order":3,"title":"Determine moles of Ca","description":"Using the $1:5$ stoichiometric ratio, we find the number of moles of calcium required by multiplying the moles of $\\text{V}_2\\text{O}_5$ by $5$."},{"type":"text","order":4,"title":"Calculate mass of Ca","content":"Finally, we convert the moles of calcium to mass using calcium's relative atomic mass ($A_r = 40.1$):\n\n$$\\begin{aligned} \\text{mass of Ca} &= n(\\text{Ca}) \\times A_r(\\text{Ca}) \\\\ &= 1.25\\text{ mol} \\times 40.1\\text{ g mol}^{-1} \\\\ &= 50.1\\text{ g} \\end{aligned}$$\n\nThus, the minimum mass of calcium required is **$50.1\\text{ g}$**.","calculator":[{"gdc":{"expression":"1.25 * 40.1","instructions":"Enter 1.25 * 40.1 in the calculation screen and press EXE."},"ti84":{"expression":"1.25 * 40.1","instructions":"Enter 1.25 multiplied by 40.1 on the home screen and press ENTER."},"desmos":{"expression":"1.25 * 40.1","instructions":"Type 1.25 * 40.1 in the input line."},"description":"Calculate the mass of calcium by multiplying the moles of Ca by its atomic mass."}],"highlights":[{"text":"\\text{Ca} = 40.1","location":"specification"}]}]},{"id":"mass_ratio_proportion","label":"Mass Ratio Method","steps":[{"type":"text","order":0,"title":"Write the balanced equation","content":"To find the relationship between vanadium(V) oxide and calcium, we first need a balanced chemical equation. Since vanadium(V) oxide, $\\text{V}_2\\text{O}_5$, is reduced to vanadium metal ($\\text{V}$), the oxygen is transferred to calcium, forming calcium oxide ($\\text{CaO}$):\n\n$$\\text{V}_2\\text{O}_5 + 5\\text{Ca} \\rightarrow 2\\text{V} + 5\\text{CaO}$$\n\nThis tells us that $1\\text{ mol}$ of $\\text{V}_2\\text{O}_5$ reacts completely with $5\\text{ mol}$ of $\\text{Ca}$."},{"type":"text","order":1,"title":"Determine the reacting masses","content":"Next, we calculate the reacting masses for the stoichiometric ratio shown in the balanced equation.\n\nFirst, calculate the molar mass ($M_r$) of vanadium(V) oxide, $\\text{V}_2\\text{O}_5$, using the provided relative atomic masses ($A_r$):\n\n$$M_r(\\text{V}_2\\text{O}_5) = (2 \\times 50.9) + (5 \\times 16.0) = 101.8 + 80.0 = 181.8\\text{ g mol}^{-1}$$\n\nThus, $1\\text{ mol}$ of $\\text{V}_2\\text{O}_5$ has a mass of $181.8\\text{ g}$.\n\nNow, calculate the mass of $5\\text{ mol}$ of calcium ($A_r(\\text{Ca}) = 40.1$):\n\n$$\\text{Mass of } 5\\text{Ca} = 5 \\times 40.1 = 200.5\\text{ g}$$\n\nAccording to the equation, **$181.8\\text{ g}$ of $\\text{V}_2\\text{O}_5$ reacts with $200.5\\text{ g}$ of $\\text{Ca}$**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Mass of } \\text{V}_2\\text{O}_5 \\rightarrow \\text{Mass of } \\text{Ca}$"},{"latex":"$$\\htmlData{anchor=std-v}{\\color{@sky}{181.8\\text{ g}}} \\rightarrow \\htmlData{anchor=std-ca}{\\color{@amber}{200.5\\text{ g}}}$"},{"latex":"$$\\htmlData{anchor=act-v}{\\color{@sky}{45.45\\text{ g}}} \\rightarrow \\htmlData{anchor=act-ca}{\\color{@amber}{x}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"std-v"},{"color":"sky","style":"text-color","anchor":"act-v"},{"color":"amber","style":"text-color","anchor":"std-ca"}],"connections":[{"to":"act-v","from":"std-v","color":"sky","label":"ratio","style":"solid"},{"to":"act-ca","from":"std-ca","color":"amber","label":"ratio","style":"solid"}]},"order":2,"title":"Set up a direct proportion","description":"We can set up a direct proportion (mass ratio) to determine the required mass of calcium ($x$) to react with $45.45\\text{ g}$ of $\\text{V}_2\\text{O}_5$."},{"type":"text","order":3,"title":"Calculate the mass of calcium","content":"Using the direct proportion ratio, we calculate the required mass of calcium:\n\n$$\\frac{x}{45.45} = \\frac{200.5}{181.8}$$\n\n$$x = 45.45 \\times \\frac{200.5}{181.8}$$\n\nLet's evaluate this calculation:\n\n$$x = 50.125\\text{ g}$$\n\nRounding to an appropriate number of significant figures (typically 3 significant figures to match the given values), the mass of calcium required is **$50.1\\text{ g}$**.","calculator":[{"gdc":{"expression":"45.45 * (200.5 / 181.8)","instructions":"Type '45.45 * (200.5 / 181.8)' and press ENTER."},"ti84":{"expression":"45.45 * (200.5 / 181.8)","instructions":"Type '45.45 * (200.5 / 181.8)' and press ENTER."},"desmos":{"expression":"45.45 \\cdot \\frac{200.5}{181.8}","instructions":"Type '45.45 * (200.5 / 181.8)' into the expression line to find the result."},"description":"Calculate the mass of calcium directly by entering the proportion formula."}]}]},{"id":"dimensional_analysis","label":"Dimensional Analysis","steps":[{"type":"text","order":0,"title":"Write the balanced chemical equation","content":"First, we need to determine the stoichiometric relationship between vanadium(V) oxide, $\\text{V}_2\\text{O}_5$, and calcium, $\\text{Ca}$. \n\nVanadium(V) oxide reacts with calcium to produce vanadium metal ($\\text{V}$) and calcium oxide ($\\text{CaO}$):\n\n$$\\text{V}_2\\text{O}_5 + 5\\text{Ca} \\rightarrow 2\\text{V} + 5\\text{CaO}$$\n\nThis balanced equation shows that $1\\text{ mol}$ of $\\text{V}_2\\text{O}_5$ reacts with $5\\text{ mol}$ of $\\text{Ca}$. This gives us our stoichiometric conversion factor:\n\n$$\\frac{5\\text{ mol Ca}}{1\\text{ mol V}_2\\text{O}_5}$$","highlights":[{"text":"Vanadium(V) oxide, \$\\text{V}_2\\text{O}_5\$, is reduced to vanadium metal by heating with calcium.","location":"specification"}]},{"type":"text","order":1,"title":"Calculate molar masses","content":"Next, we calculate the molar masses using the relative atomic masses provided ($\\text{V} = 50.9$, $\\text{Ca} = 40.1$, $\\text{O} = 16.0$):\n\n* **Molar mass of $\\text{V}_2\\text{O}_5$**:\n $$M_r(\\text{V}_2\\text{O}_5) = (2 \\times 50.9) + (5 \\times 16.0) = 181.8\\text{ g mol}^{-1}$$\n This gives the conversion factor: $\\frac{1\\text{ mol V}_2\\text{O}_5}{181.8\\text{ g V}_2\\text{O}_5}$\n\n* **Molar mass of $\\text{Ca}$**:\n $$M_r(\\text{Ca}) = 40.1\\text{ g mol}^{-1}$$\n This gives the conversion factor: $\\frac{40.1\\text{ g Ca}}{1\\text{ mol Ca}}$","highlights":[{"text":"\$\\text{V} = 50.9\$, \$\\text{Ca} = 40.1\$, \$\\text{O} = 16.0\$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"We start with the given mass of vanadium(V) oxide ($45.45\\text{ g}$) and apply our conversion factors in a continuous chain to cancel out units and find the mass of calcium:"},{"latex":"$$$\\text{mass of Ca} = \\htmlData{anchor=start-mass}{\\color{@sky}{45.45\\text{ g V}_2\\text{O}_5}} \\times \\htmlData{anchor=mol-v-conv}{\\color{@amber}{\\frac{1\\text{ mol V}_2\\text{O}_5}{181.8\\text{ g V}_2\\text{O}_5}}} \\times \\htmlData{anchor=ratio-conv}{\\color{@purple}{\\frac{5\\text{ mol Ca}}{1\\text{ mol V}_2\\text{O}_5}}} \\times \\htmlData{anchor=mol-ca-conv}{\\color{@teal}{\\frac{40.1\\text{ g Ca}}{1\\text{ mol Ca}}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"start-mass"},{"color":"amber","style":"text-color","anchor":"mol-v-conv"},{"color":"purple","style":"text-color","anchor":"ratio-conv"},{"color":"teal","style":"text-color","anchor":"mol-ca-conv"}],"connections":[{"to":"mol-v-conv","from":"start-mass","color":"sky","label":"convert to moles","style":"solid"},{"to":"ratio-conv","from":"mol-v-conv","color":"amber","label":"apply mole ratio","style":"solid"},{"to":"mol-ca-conv","from":"ratio-conv","color":"purple","label":"convert to mass","style":"solid"}]},"order":2,"title":"Set up dimensional analysis","description":"Each step in the multiplication cancels the previous unit, leaving us with the target unit (g Ca)."},{"type":"text","order":3,"title":"Calculate the final mass","content":"Now, we carry out the calculation:\n\n$$\\text{mass of Ca} = 45.45 \\times \\frac{1}{181.8} \\times 5 \\times 40.1 = 50.1\\text{ g}$$\n\nThus, the minimum mass of calcium required is **$50.1\\text{ g}$**.","calculator":[{"gdc":{"expression":"45.45 * (1 / 181.8) * 5 * 40.1","instructions":"Enter the continuous multiplication directly:\n45.45 * (1 / 181.8) * 5 * 40.1"},"ti84":{"expression":"45.45 * (1 / 181.8) * 5 * 40.1","instructions":"Enter the expression into the home screen:\n45.45 * (1 / 181.8) * 5 * 40.1"},"desmos":{"expression":"45.45 * (1 / 181.8) * 5 * 40.1","instructions":"Type the continuous product into the cell:\n45.45 * (1 / 181.8) * 5 * 40.1"},"description":"Perform the dimensional analysis calculation in one step on your calculator."}]}]}],"generatedAt":"2026-06-11T09:25:37.558Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"d86c281f-7463-466f-9620-23d748d224bd:stepwise_mole_method","label":"Stepwise Mole Calculation","steps":[{"type":"text","order":0,"title":"Write the balanced equation","content":"To find the relationship between vanadium(V) oxide ($\\text{V}_2\\text{O}_5$) and calcium ($\\text{Ca}$), we must first write a balanced chemical equation for the reduction reaction:\n\n$$\\text{V}_2\\text{O}_5 + 5\\text{Ca} \\rightarrow 2\\text{V} + 5\\text{CaO}$$\n\nThis shows us that the stoichiometric ratio between $\\text{V}_2\\text{O}_5$ and $\\text{Ca}$ is **$1:5$**.","highlights":[{"text":"Vanadium(V) oxide, \$\\text{V}_2\\text{O}_5\$, is reduced to vanadium metal by heating with calcium.","location":"specification"}]},{"type":"text","order":1,"title":"Calculate molar mass of V2O5","content":"Next, we calculate the molar mass ($M_r$) of vanadium(V) oxide, $\\text{V}_2\\text{O}_5$, using the given relative atomic masses ($\\text{V} = 50.9$ and $\\text{O} = 16.0$):\n\n$$\\begin{aligned} M_r(\\text{V}_2\\text{O}_5) &= (2 \\times 50.9) + (5 \\times 16.0) \\\\ &= 101.8 + 80.0 \\\\ &= 181.8\\text{ g mol}^{-1} \\end{aligned}$$","highlights":[{"text":"\\text{V} = 50.9","location":"specification"},{"text":"\\text{O} = 16.0","location":"specification"}]},{"type":"text","order":2,"title":"Calculate moles of V2O5","content":"Now, we calculate the chemical amount in moles of $\\text{V}_2\\text{O}_5$ present in $45.45\\text{ g}$ of the substance using the formula:\n\n$$\\text{moles} = \\frac{\\text{mass}}{\\text{molar mass}}$$\n\n$$\\begin{aligned} n(\\text{V}_2\\text{O}_5) &= \\frac{45.45\\text{ g}}{181.8\\text{ g mol}^{-1}} \\\\ &= 0.250\\text{ mol} \\end{aligned}$$","calculator":[{"gdc":{"expression":"45.45 / 181.8","instructions":"Enter 45.45 / 181.8 in the calculation screen and press EXE."},"ti84":{"expression":"45.45 / 181.8","instructions":"Enter 45.45 divided by 181.8 on the home screen and press ENTER."},"desmos":{"expression":"45.45 / 181.8","instructions":"Type 45.45 / 181.8 in the input line."},"description":"Calculate the moles of vanadium(V) oxide by dividing the mass by the molar mass."}],"highlights":[{"text":"45.45\\text{ g}","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{molar ratio of } \\text{V}_2\\text{O}_5 : \\text{Ca} = \\htmlData{anchor=ratio-v}{\\color{@sky}{1}} : \\htmlData{anchor=ratio-ca}{\\color{@amber}{5}}$"},{"latex":"$$n(\\text{Ca}) = \\htmlData{anchor=factor-ca}{\\color{@amber}{5}} \\times \\htmlData{anchor=moles-v}{\\color{@sky}{0.250}}\\text{ mol} = \\htmlData{anchor=moles-ca}{\\color{@purple}{1.25}}\\text{ mol}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"ratio-v"},{"color":"amber","style":"text-color","anchor":"ratio-ca"},{"color":"sky","style":"text-color","anchor":"moles-v"},{"color":"purple","style":"text-color","anchor":"moles-ca"}],"connections":[{"to":"factor-ca","from":"ratio-ca","color":"amber","label":"multiply factor","style":"solid"},{"to":"moles-ca","from":"moles-v","color":"purple","label":null,"style":"dashed"}]},"order":3,"title":"Determine moles of Ca","description":"Using the $1:5$ stoichiometric ratio, we find the number of moles of calcium required by multiplying the moles of $\\text{V}_2\\text{O}_5$ by $5$."},{"type":"text","order":4,"title":"Calculate mass of Ca","content":"Finally, we convert the moles of calcium to mass using calcium's relative atomic mass ($A_r = 40.1$):\n\n$$\\begin{aligned} \\text{mass of Ca} &= n(\\text{Ca}) \\times A_r(\\text{Ca}) \\\\ &= 1.25\\text{ mol} \\times 40.1\\text{ g mol}^{-1} \\\\ &= 50.1\\text{ g} \\end{aligned}$$\n\nThus, the minimum mass of calcium required is **$50.1\\text{ g}$**.","calculator":[{"gdc":{"expression":"1.25 * 40.1","instructions":"Enter 1.25 * 40.1 in the calculation screen and press EXE."},"ti84":{"expression":"1.25 * 40.1","instructions":"Enter 1.25 multiplied by 40.1 on the home screen and press ENTER."},"desmos":{"expression":"1.25 * 40.1","instructions":"Type 1.25 * 40.1 in the input line."},"description":"Calculate the mass of calcium by multiplying the moles of Ca by its atomic mass."}],"highlights":[{"text":"\\text{Ca} = 40.1","location":"specification"}]}]},{"id":"d86c281f-7463-466f-9620-23d748d224bd:mass_ratio_proportion","label":"Mass Ratio Method","steps":[{"type":"text","order":0,"title":"Write the balanced equation","content":"To find the relationship between vanadium(V) oxide and calcium, we first need a balanced chemical equation. Since vanadium(V) oxide, $\\text{V}_2\\text{O}_5$, is reduced to vanadium metal ($\\text{V}$), the oxygen is transferred to calcium, forming calcium oxide ($\\text{CaO}$):\n\n$$\\text{V}_2\\text{O}_5 + 5\\text{Ca} \\rightarrow 2\\text{V} + 5\\text{CaO}$$\n\nThis tells us that $1\\text{ mol}$ of $\\text{V}_2\\text{O}_5$ reacts completely with $5\\text{ mol}$ of $\\text{Ca}$."},{"type":"text","order":1,"title":"Determine the reacting masses","content":"Next, we calculate the reacting masses for the stoichiometric ratio shown in the balanced equation.\n\nFirst, calculate the molar mass ($M_r$) of vanadium(V) oxide, $\\text{V}_2\\text{O}_5$, using the provided relative atomic masses ($A_r$):\n\n$$M_r(\\text{V}_2\\text{O}_5) = (2 \\times 50.9) + (5 \\times 16.0) = 101.8 + 80.0 = 181.8\\text{ g mol}^{-1}$$\n\nThus, $1\\text{ mol}$ of $\\text{V}_2\\text{O}_5$ has a mass of $181.8\\text{ g}$.\n\nNow, calculate the mass of $5\\text{ mol}$ of calcium ($A_r(\\text{Ca}) = 40.1$):\n\n$$\\text{Mass of } 5\\text{Ca} = 5 \\times 40.1 = 200.5\\text{ g}$$\n\nAccording to the equation, **$181.8\\text{ g}$ of $\\text{V}_2\\text{O}_5$ reacts with $200.5\\text{ g}$ of $\\text{Ca}$**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Mass of } \\text{V}_2\\text{O}_5 \\rightarrow \\text{Mass of } \\text{Ca}$"},{"latex":"$$\\htmlData{anchor=std-v}{\\color{@sky}{181.8\\text{ g}}} \\rightarrow \\htmlData{anchor=std-ca}{\\color{@amber}{200.5\\text{ g}}}$"},{"latex":"$$\\htmlData{anchor=act-v}{\\color{@sky}{45.45\\text{ g}}} \\rightarrow \\htmlData{anchor=act-ca}{\\color{@amber}{x}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"std-v"},{"color":"sky","style":"text-color","anchor":"act-v"},{"color":"amber","style":"text-color","anchor":"std-ca"}],"connections":[{"to":"act-v","from":"std-v","color":"sky","label":"ratio","style":"solid"},{"to":"act-ca","from":"std-ca","color":"amber","label":"ratio","style":"solid"}]},"order":2,"title":"Set up a direct proportion","description":"We can set up a direct proportion (mass ratio) to determine the required mass of calcium ($x$) to react with $45.45\\text{ g}$ of $\\text{V}_2\\text{O}_5$."},{"type":"text","order":3,"title":"Calculate the mass of calcium","content":"Using the direct proportion ratio, we calculate the required mass of calcium:\n\n$$\\frac{x}{45.45} = \\frac{200.5}{181.8}$$\n\n$$x = 45.45 \\times \\frac{200.5}{181.8}$$\n\nLet's evaluate this calculation:\n\n$$x = 50.125\\text{ g}$$\n\nRounding to an appropriate number of significant figures (typically 3 significant figures to match the given values), the mass of calcium required is **$50.1\\text{ g}$**.","calculator":[{"gdc":{"expression":"45.45 * (200.5 / 181.8)","instructions":"Type '45.45 * (200.5 / 181.8)' and press ENTER."},"ti84":{"expression":"45.45 * (200.5 / 181.8)","instructions":"Type '45.45 * (200.5 / 181.8)' and press ENTER."},"desmos":{"expression":"45.45 \\cdot \\frac{200.5}{181.8}","instructions":"Type '45.45 * (200.5 / 181.8)' into the expression line to find the result."},"description":"Calculate the mass of calcium directly by entering the proportion formula."}]}]},{"id":"d86c281f-7463-466f-9620-23d748d224bd:dimensional_analysis","label":"Dimensional Analysis","steps":[{"type":"text","order":0,"title":"Write the balanced chemical equation","content":"First, we need to determine the stoichiometric relationship between vanadium(V) oxide, $\\text{V}_2\\text{O}_5$, and calcium, $\\text{Ca}$. \n\nVanadium(V) oxide reacts with calcium to produce vanadium metal ($\\text{V}$) and calcium oxide ($\\text{CaO}$):\n\n$$\\text{V}_2\\text{O}_5 + 5\\text{Ca} \\rightarrow 2\\text{V} + 5\\text{CaO}$$\n\nThis balanced equation shows that $1\\text{ mol}$ of $\\text{V}_2\\text{O}_5$ reacts with $5\\text{ mol}$ of $\\text{Ca}$. This gives us our stoichiometric conversion factor:\n\n$$\\frac{5\\text{ mol Ca}}{1\\text{ mol V}_2\\text{O}_5}$$","highlights":[{"text":"Vanadium(V) oxide, \$\\text{V}_2\\text{O}_5\$, is reduced to vanadium metal by heating with calcium.","location":"specification"}]},{"type":"text","order":1,"title":"Calculate molar masses","content":"Next, we calculate the molar masses using the relative atomic masses provided ($\\text{V} = 50.9$, $\\text{Ca} = 40.1$, $\\text{O} = 16.0$):\n\n* **Molar mass of $\\text{V}_2\\text{O}_5$**:\n $$M_r(\\text{V}_2\\text{O}_5) = (2 \\times 50.9) + (5 \\times 16.0) = 181.8\\text{ g mol}^{-1}$$\n This gives the conversion factor: $\\frac{1\\text{ mol V}_2\\text{O}_5}{181.8\\text{ g V}_2\\text{O}_5}$\n\n* **Molar mass of $\\text{Ca}$**:\n $$M_r(\\text{Ca}) = 40.1\\text{ g mol}^{-1}$$\n This gives the conversion factor: $\\frac{40.1\\text{ g Ca}}{1\\text{ mol Ca}}$","highlights":[{"text":"\$\\text{V} = 50.9\$, \$\\text{Ca} = 40.1\$, \$\\text{O} = 16.0\$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"We start with the given mass of vanadium(V) oxide ($45.45\\text{ g}$) and apply our conversion factors in a continuous chain to cancel out units and find the mass of calcium:"},{"latex":"$$$\\text{mass of Ca} = \\htmlData{anchor=start-mass}{\\color{@sky}{45.45\\text{ g V}_2\\text{O}_5}} \\times \\htmlData{anchor=mol-v-conv}{\\color{@amber}{\\frac{1\\text{ mol V}_2\\text{O}_5}{181.8\\text{ g V}_2\\text{O}_5}}} \\times \\htmlData{anchor=ratio-conv}{\\color{@purple}{\\frac{5\\text{ mol Ca}}{1\\text{ mol V}_2\\text{O}_5}}} \\times \\htmlData{anchor=mol-ca-conv}{\\color{@teal}{\\frac{40.1\\text{ g Ca}}{1\\text{ mol Ca}}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"start-mass"},{"color":"amber","style":"text-color","anchor":"mol-v-conv"},{"color":"purple","style":"text-color","anchor":"ratio-conv"},{"color":"teal","style":"text-color","anchor":"mol-ca-conv"}],"connections":[{"to":"mol-v-conv","from":"start-mass","color":"sky","label":"convert to moles","style":"solid"},{"to":"ratio-conv","from":"mol-v-conv","color":"amber","label":"apply mole ratio","style":"solid"},{"to":"mol-ca-conv","from":"ratio-conv","color":"purple","label":"convert to mass","style":"solid"}]},"order":2,"title":"Set up dimensional analysis","description":"Each step in the multiplication cancels the previous unit, leaving us with the target unit (g Ca)."},{"type":"text","order":3,"title":"Calculate the final mass","content":"Now, we carry out the calculation:\n\n$$\\text{mass of Ca} = 45.45 \\times \\frac{1}{181.8} \\times 5 \\times 40.1 = 50.1\\text{ g}$$\n\nThus, the minimum mass of calcium required is **$50.1\\text{ g}$**.","calculator":[{"gdc":{"expression":"45.45 * (1 / 181.8) * 5 * 40.1","instructions":"Enter the continuous multiplication directly:\n45.45 * (1 / 181.8) * 5 * 40.1"},"ti84":{"expression":"45.45 * (1 / 181.8) * 5 * 40.1","instructions":"Enter the expression into the home screen:\n45.45 * (1 / 181.8) * 5 * 40.1"},"desmos":{"expression":"45.45 * (1 / 181.8) * 5 * 40.1","instructions":"Type the continuous product into the cell:\n45.45 * (1 / 181.8) * 5 * 40.1"},"description":"Perform the dimensional analysis calculation in one step on your calculator."}]}]}],"generatedAt":"2026-06-11T09:25:37.621Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"682d2613-bdd3-4b5d-9aa3-1a5abbed6da8","revisionId":"30f86aa3-8401-4d8c-90e9-69411b9c13e2","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"4c69c009-0121-4f68-b9e9-f2e4bfb79a5b","label":null,"content":"Compound \$\\textbf{X}\$ is a chloride of sulfur that is a liquid at room temperature and pressure and has a boiling point of \$59^\\circ\\text{C}\$.\n\nWhen \$0.450\\text{ g}\$ of \$\\textbf{X}\$ is heated to \$95^\\circ\\text{C}\$ at \$9.80 \\times 10^4\\text{ Pa}\$, \$136\\text{ cm}^3\$ of gas is produced.\n\nDetermine the molecular formula of compound \$\\textbf{X}\$.\n\nShow all your working.","metadata":null,"markscheme":"- Converts temperature to $368\\text{ K}$ and volume to $1.36 \\times 10^{-4}\\text{ m}^3$ M1\n- Calculates amount of gas using $pV = nRT$: $n = \\frac{9.80 \\times 10^4 \\times 1.36 \\times 10^{-4}}{8.31 \\times 368} \\approx 4.36 \\times 10^{-3}\\text{ mol}$ (accept $4.35 \\times 10^{-3}$ to $4.36 \\times 10^{-3}$ mol) M1\n- Calculates molar mass: $M = \\frac{0.450}{4.358 \\times 10^{-3}} \\approx 103.3\\text{ g mol}^{-1}$ (accept $103.1$ to $103.5\\text{ g mol}^{-1}$) M1\n- Subtracts sulfur's mass ($103.3 - 32.1 = 71.2$) and divides by chlorine's mass ($\\frac{71.2}{35.5} \\approx 2$), identifying the molecular formula as $\\text{SCl}_2$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"twostep","label":"Calculating Moles First","steps":[{"type":"text","order":0,"title":"Convert units to SI","content":"Before using the ideal gas equation, we must ensure all quantities are in SI units:\n- **Temperature ($T$):** Convert from Celsius to Kelvin:\n $$T = 95 + 273 = 368\\text{ K}$$\n- **Volume ($V$):** Convert from cubic centimeters ($\\text{cm}^3$) to cubic meters ($\\text{m}^3$):\n $$V = 136\\text{ cm}^3 = 136 \\times 10^{-6}\\text{ m}^3 = 1.36 \\times 10^{-4}\\text{ m}^3$$\n- **Pressure ($p$):** Already in Pascals:\n $$p = 9.80 \\times 10^4\\text{ Pa}$$\n\nConverting both values correctly earns you the first mark M1.","highlights":[{"text":"95°C","location":"specification"},{"text":"136 cm3","location":"specification"}]},{"type":"text","order":1,"title":"Calculate moles of gas","content":"Using the ideal gas equation from your data sheet:\n\n$$pV = nRT$$\n\nRearranging for the number of moles ($n$):\n\n$$n = \\frac{pV}{RT}$$\n\nSubstitute our values into the formula using the gas constant $R = 8.31\\text{ J K}^{-1}\\text{ mol}^{-1}$:\n\n$$n = \\frac{9.80 \\times 10^4 \\times 1.36 \\times 10^{-4}}{8.31 \\times 368} \\approx 4.36 \\times 10^{-3}\\text{ mol}$$\n\nThis calculation earns you the second mark M1.","calculator":[{"gdc":{"expression":"(9.80*10^(4) * 1.36*10^(-4)) / (8.31 * 368)","instructions":"Use the fraction template and enter:\nNumerator: 9.80E4 * 1.36E-4\nDenominator: 8.31 * 368"},"ti84":{"expression":"(9.80*10^(4) * 1.36*10^(-4)) / (8.31 * 368)","instructions":"Enter the numerator inside parentheses, then divide by the denominator inside parentheses:\n(9.80*10^(4) * 1.36*10^(-4)) / (8.31 * 368)"},"desmos":{"expression":"\\frac{9.80 \\cdot 10^{4} \\cdot 1.36 \\cdot 10^{-4}}{8.31 \\cdot 368}","instructions":"Type the division helper directly:\n(9.80 * 10^4 * 1.36 * 10^-4) / (8.31 * 368)"},"description":"Perform the division on a scientific or graphing calculator to find the number of moles."}]},{"type":"text","order":2,"title":"Determine the molar mass","content":"Now, we relate the mass of the sample to the calculated number of moles to find the molar mass ($M$):\n\n$$M = \\frac{\\text{mass}}{n} = \\frac{0.450\\text{ g}}{4.358 \\times 10^{-3}\\text{ mol}} \\approx 103.3\\text{ g mol}^{-1}$$\n\nFinding this molar mass earns the third mark M1.","calculator":[{"gdc":{"expression":"0.450 / Ans","instructions":"Divide 0.450 by the previous answer:\n0.450 / Ans"},"ti84":{"expression":"0.450 / Ans","instructions":"Divide 0.450 by the previous answer:\n0.450 / Ans"},"desmos":{"expression":"0.450 / 0.004358485","instructions":"Calculate the division using the exact decimal:\n0.450 / 0.004358485"},"description":"Divide the given mass by the precise value of moles stored in your calculator's memory."}],"highlights":[{"text":"0.450 g","location":"specification"}]},{"type":"text","order":3,"title":"Deduce the molecular formula","content":"Compound $\\textbf{X}$ is a chloride of sulfur ($\\text{S}_x\\text{Cl}_y$). Let's use standard atomic masses from the periodic table ($M(\\text{S}) = 32.1\\text{ g mol}^{-1}$ and $M(\\text{Cl}) = 35.5\\text{ g mol}^{-1}$):\n\nSubtract the mass of one sulfur atom from the molar mass:\n\n$$103.3 - 32.1 = 71.2\\text{ g mol}^{-1}$$\n\nDivide the remaining mass by the atomic mass of chlorine:\n\n$$\\frac{71.2}{35.5} \\approx 2$$\n\nThis indicates there is $1$ sulfur atom and $2$ chlorine atoms per molecule.\n\nTherefore, the molecular formula of compound $\\textbf{X}$ is **$\\text{SCl}_2$** (sulfur dichloride).\n\nCorrectly identifying the formula with working earns the final mark A1."}]},{"id":"onestep","label":"Direct Molar Mass Formula","steps":[{"type":"text","order":0,"title":"Identify given quantities","content":"To find the molar mass of compound $\\textbf{X}$ directly, we first need to identify the given quantities and convert them into standard SI units for the ideal gas equation:\n\n* **Mass ($m$):** $0.450\\text{ g}$\n* **Temperature ($T$):** $95^\\circ\\text{C}$\n* **Pressure ($p$):** $9.80 \\times 10^4\\text{ Pa}$\n* **Volume ($V$):** $136\\text{ cm}^3$\n* **Gas constant ($R$):** $8.31\\text{ J K}^{-1}\\text{ mol}^{-1}$","highlights":[{"text":"0.450 g","location":"specification"},{"text":"95°C","location":"specification"},{"text":"9.80 × 10^4 Pa","location":"specification"},{"text":"136 cm^3","location":"specification"}]},{"type":"text","order":1,"title":"Convert units to SI","content":"Convert temperature ($T$) to Kelvin and volume ($V$) to cubic meters ($m^3$) so they align with the SI units of the gas constant $R$:\n\n* **Temperature:**\n $$T = 95 + 273 = 368\\text{ K}$$\n\n* **Volume:**\n Since $1\\text{ m}^3 = 10^6\\text{ cm}^3$, we divide by $10^6$:\n $$V = 136 \\times 10^{-6}\\text{ m}^3 = 1.36 \\times 10^{-4}\\text{ m}^3$$\n\nThese conversions are essential to earn your first mark (**M1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$M = \\frac{\\htmlData{anchor=mass}{\\color{@sky}{m}}R\\htmlData{anchor=temp}{\\color{@amber}{T}}}{\\htmlData{anchor=pres}{\\color{@purple}{p}}\\htmlData{anchor=vol}{\\color{@green}{V}}}$"},{"latex":"$$M = \\frac{\\htmlData{anchor=mass-sub}{\\color{@sky}{0.450}} \\times 8.31 \\times \\htmlData{anchor=temp-sub}{\\color{@amber}{368}}}{\\htmlData{anchor=pres-sub}{\\color{@purple}{9.80 \\times 10^4}} \\times \\htmlData{anchor=vol-sub}{\\color{@green}{1.36 \\times 10^{-4}}}}$"},{"latex":"$$M \\approx 103.3\\text{ g mol}^{-1}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mass"},{"color":"amber","style":"text-color","anchor":"temp"},{"color":"purple","style":"text-color","anchor":"pres"},{"color":"green","style":"text-color","anchor":"vol"}],"connections":[{"to":"mass-sub","from":"mass","color":"sky","label":"mass in g","style":"solid"},{"to":"temp-sub","from":"temp","color":"amber","label":"temperature in K","style":"solid"},{"to":"pres-sub","from":"pres","color":"purple","label":"pressure in Pa","style":"solid"},{"to":"vol-sub","from":"vol","color":"green","label":"volume in m³","style":"solid"}]},"order":2,"title":"Calculate molar mass directly","description":"We rearrange the ideal gas equation $pV = nRT$ by substituting $n = \\frac{m}{M}$. This gives $pV = \\frac{mRT}{M}$, which rearranges directly to the molar mass formula $M = \\frac{mRT}{pV}$."},{"type":"text","order":3,"title":"Perform calculation using GDC","content":"To find the numerical value of $M$, evaluate the fraction carefully using your calculator. This step corresponds to earning **M2** and **M3**.","calculator":[{"gdc":{"expression":"(0.450 * 8.31 * 368) / (9.80e4 * 1.36e-4)","instructions":"Enter the calculation using the division operator, ensuring the standard notation is correct:"},"ti84":{"expression":"(0.450 * 8.31 * 368) / (9.80E4 * 1.36E-4)","instructions":"Enter the formula in your main screen using parentheses to keep the numerator and denominator separated:"},"desmos":{"expression":"\\frac{0.450 \\cdot 8.31 \\cdot 368}{9.80 \\cdot 10^4 \\cdot 1.36 \\cdot 10^{-4}}","instructions":"Type the division as a fraction to visually align the numerator and denominator:"},"description":"Calculate the direct molar mass of the compound"}]},{"type":"text","order":4,"title":"Determine the molecular formula","content":"We are given that compound $\\textbf{X}$ is a chloride of sulfur. Let's find how many chlorine atoms are present per sulfur atom:\n\n1. **Molar mass of sulfur (S):** $32.1\\text{ g mol}^{-1}$\n2. **Molar mass of chlorine (Cl):** $35.5\\text{ g mol}^{-1}$\n\nSubtract the mass of one sulfur atom from the total molar mass:\n\n$$103.3 - 32.1 = 71.2\\text{ g mol}^{-1}$$\n\nDivide the remaining mass by the molar mass of chlorine to find the number of chlorine atoms:\n\n$$\\text{Number of Cl atoms} = \\frac{71.2}{35.5} \\approx 2$$\n\nThus, there are 2 chlorine atoms for every 1 sulfur atom (**A1**).\n\n**Molecular Formula:**\n$$\\text{SCl}_2$$"}]},{"id":"backwards","label":"Hypothesis Testing","steps":[{"type":"text","order":0,"title":"Convert given conditions","content":"Before testing our hypotheses, we must convert the experimental conditions to SI units so they are compatible with the ideal gas equation, $pV = nRT$.\n\n* **Temperature ($T$):** \n $$T = 95^\\circ\\text{C} + 273 = 368\\text{ K}$$\n\n* **Pressure ($p$):** \n $$p = 9.80 \\times 10^4\\text{ Pa}$$","highlights":[{"text":"95^\\circ\\text{C}","location":"specification"},{"text":"9.80 \\times 10^4\\text{ Pa}","location":"specification"}]},{"type":"text","order":1,"title":"Define hypotheses for X","content":"We will assume potential molecular formulas for the sulfur chloride compound $\\textbf{X}$ and calculate their respective molar masses ($M$):\n\n* **Hypothesis 1:** $\\text{SCl}$\n $$M(\\text{SCl}) = 32.1 + 35.5 = 67.6\\text{ g mol}^{-1}$$\n\n* **Hypothesis 2:** $\\text{SCl}_2$\n $$M(\\text{SCl}_2) = 32.1 + 2(35.5) = 103.1\\text{ g mol}^{-1}$$\n\n* **Hypothesis 3:** $\\text{S}_2\\text{Cl}_2$\n $$M(\\text{S}_2\\text{Cl}_2) = 2(32.1) + 2(35.5) = 135.2\\text{ g mol}^{-1}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"For a sample of mass $m = 0.450\\text{ g}$, we calculate the number of moles ($n = \\frac{m}{M}$) for each candidate formula:"},{"latex":"$$$n(\\text{Hypothesis 1}) = \\frac{0.450\\text{ g}}{\\htmlData{anchor=M1}{\\color{@sky}{67.6\\text{ g mol}^{-1}}}} \\approx \\htmlData{anchor=n1}{\\color{@sky}{6.66 \\times 10^{-3}\\text{ mol}}}$$"},{"latex":"$$$n(\\text{Hypothesis 2}) = \\frac{0.450\\text{ g}}{\\htmlData{anchor=M2}{\\color{@amber}{103.1\\text{ g mol}^{-1}}}} \\approx \\htmlData{anchor=n2}{\\color{@amber}{4.36 \\times 10^{-3}\\text{ mol}}}$$"},{"latex":"$$$n(\\text{Hypothesis 3}) = \\frac{0.450\\text{ g}}{\\htmlData{anchor=M3}{\\color{@purple}{135.2\\text{ g mol}^{-1}}}} \\approx \\htmlData{anchor=n3}{\\color{@purple}{3.33 \\times 10^{-3}\\text{ mol}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"M1"},{"color":"sky","style":"text-color","anchor":"n1"},{"color":"amber","style":"text-color","anchor":"M2"},{"color":"amber","style":"text-color","anchor":"n2"},{"color":"purple","style":"text-color","anchor":"M3"},{"color":"purple","style":"text-color","anchor":"n3"}],"connections":[{"to":"n1","from":"M1","color":"sky","label":"n = m/M","style":"solid"},{"to":"n2","from":"M2","color":"amber","label":"n = m/M","style":"solid"},{"to":"n3","from":"M3","color":"purple","label":"n = m/M","style":"solid"}]},"order":2,"title":"Calculate theoretical moles","description":"Calculate the moles corresponding to $0.450\\text{ g}$ of each candidate sulfur chloride."},{"type":"text","order":3,"title":"Calculate theoretical volumes","content":"Using the ideal gas equation rearranged for volume ($V = \\frac{nRT}{p}$), where $R = 8.31\\text{ J K}^{-1}\\text{ mol}^{-1}$:\n\n* **For Hypothesis 1 ($\\text{SCl}$):**\n $$V = \\frac{6.66 \\times 10^{-3} \\times 8.31 \\times 368}{9.80 \\times 10^4} \\approx 2.08 \\times 10^{-4}\\text{ m}^3 = 208\\text{ cm}^3$$\n\n* **For Hypothesis 2 ($\\text{SCl}_2$):**\n $$V = \\frac{4.36 \\times 10^{-3} \\times 8.31 \\times 368}{9.80 \\times 10^4} \\approx 1.36 \\times 10^{-4}\\text{ m}^3 = 136\\text{ cm}^3$$\n\n* **For Hypothesis 3 ($\\text{S}_2\\text{Cl}_2$):**\n $$V = \\frac{3.33 \\times 10^{-3} \\times 8.31 \\times 368}{9.80 \\times 10^4} \\approx 1.04 \\times 10^{-4}\\text{ m}^3 = 104\\text{ cm}^3$$"},{"type":"text","order":4,"title":"Identify correct formula","content":"Comparing our computed volumes with the measured volume of $136\\text{ cm}^3$:\n\n* **Hypothesis 2** ($\\text{SCl}_2$) yields a theoretical volume of exactly $136\\text{ cm}^3$.\n\nTherefore, the molecular formula of compound $\\textbf{X}$ is **$\\text{SCl}_2$**.","highlights":[{"text":"136\\text{ cm}^3","location":"specification"}]}]}],"generatedAt":"2026-06-11T09:26:34.193Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"4c69c009-0121-4f68-b9e9-f2e4bfb79a5b:twostep","label":"Calculating Moles First","steps":[{"type":"text","order":0,"title":"Convert units to SI","content":"Before using the ideal gas equation, we must ensure all quantities are in SI units:\n- **Temperature ($T$):** Convert from Celsius to Kelvin:\n $$T = 95 + 273 = 368\\text{ K}$$\n- **Volume ($V$):** Convert from cubic centimeters ($\\text{cm}^3$) to cubic meters ($\\text{m}^3$):\n $$V = 136\\text{ cm}^3 = 136 \\times 10^{-6}\\text{ m}^3 = 1.36 \\times 10^{-4}\\text{ m}^3$$\n- **Pressure ($p$):** Already in Pascals:\n $$p = 9.80 \\times 10^4\\text{ Pa}$$\n\nConverting both values correctly earns you the first mark M1.","highlights":[{"text":"95°C","location":"specification"},{"text":"136 cm3","location":"specification"}]},{"type":"text","order":1,"title":"Calculate moles of gas","content":"Using the ideal gas equation from your data sheet:\n\n$$pV = nRT$$\n\nRearranging for the number of moles ($n$):\n\n$$n = \\frac{pV}{RT}$$\n\nSubstitute our values into the formula using the gas constant $R = 8.31\\text{ J K}^{-1}\\text{ mol}^{-1}$:\n\n$$n = \\frac{9.80 \\times 10^4 \\times 1.36 \\times 10^{-4}}{8.31 \\times 368} \\approx 4.36 \\times 10^{-3}\\text{ mol}$$\n\nThis calculation earns you the second mark M1.","calculator":[{"gdc":{"expression":"(9.80*10^(4) * 1.36*10^(-4)) / (8.31 * 368)","instructions":"Use the fraction template and enter:\nNumerator: 9.80E4 * 1.36E-4\nDenominator: 8.31 * 368"},"ti84":{"expression":"(9.80*10^(4) * 1.36*10^(-4)) / (8.31 * 368)","instructions":"Enter the numerator inside parentheses, then divide by the denominator inside parentheses:\n(9.80*10^(4) * 1.36*10^(-4)) / (8.31 * 368)"},"desmos":{"expression":"\\frac{9.80 \\cdot 10^{4} \\cdot 1.36 \\cdot 10^{-4}}{8.31 \\cdot 368}","instructions":"Type the division helper directly:\n(9.80 * 10^4 * 1.36 * 10^-4) / (8.31 * 368)"},"description":"Perform the division on a scientific or graphing calculator to find the number of moles."}]},{"type":"text","order":2,"title":"Determine the molar mass","content":"Now, we relate the mass of the sample to the calculated number of moles to find the molar mass ($M$):\n\n$$M = \\frac{\\text{mass}}{n} = \\frac{0.450\\text{ g}}{4.358 \\times 10^{-3}\\text{ mol}} \\approx 103.3\\text{ g mol}^{-1}$$\n\nFinding this molar mass earns the third mark M1.","calculator":[{"gdc":{"expression":"0.450 / Ans","instructions":"Divide 0.450 by the previous answer:\n0.450 / Ans"},"ti84":{"expression":"0.450 / Ans","instructions":"Divide 0.450 by the previous answer:\n0.450 / Ans"},"desmos":{"expression":"0.450 / 0.004358485","instructions":"Calculate the division using the exact decimal:\n0.450 / 0.004358485"},"description":"Divide the given mass by the precise value of moles stored in your calculator's memory."}],"highlights":[{"text":"0.450 g","location":"specification"}]},{"type":"text","order":3,"title":"Deduce the molecular formula","content":"Compound $\\textbf{X}$ is a chloride of sulfur ($\\text{S}_x\\text{Cl}_y$). Let's use standard atomic masses from the periodic table ($M(\\text{S}) = 32.1\\text{ g mol}^{-1}$ and $M(\\text{Cl}) = 35.5\\text{ g mol}^{-1}$):\n\nSubtract the mass of one sulfur atom from the molar mass:\n\n$$103.3 - 32.1 = 71.2\\text{ g mol}^{-1}$$\n\nDivide the remaining mass by the atomic mass of chlorine:\n\n$$\\frac{71.2}{35.5} \\approx 2$$\n\nThis indicates there is $1$ sulfur atom and $2$ chlorine atoms per molecule.\n\nTherefore, the molecular formula of compound $\\textbf{X}$ is **$\\text{SCl}_2$** (sulfur dichloride).\n\nCorrectly identifying the formula with working earns the final mark A1."}]},{"id":"4c69c009-0121-4f68-b9e9-f2e4bfb79a5b:onestep","label":"Direct Molar Mass Formula","steps":[{"type":"text","order":0,"title":"Identify given quantities","content":"To find the molar mass of compound $\\textbf{X}$ directly, we first need to identify the given quantities and convert them into standard SI units for the ideal gas equation:\n\n* **Mass ($m$):** $0.450\\text{ g}$\n* **Temperature ($T$):** $95^\\circ\\text{C}$\n* **Pressure ($p$):** $9.80 \\times 10^4\\text{ Pa}$\n* **Volume ($V$):** $136\\text{ cm}^3$\n* **Gas constant ($R$):** $8.31\\text{ J K}^{-1}\\text{ mol}^{-1}$","highlights":[{"text":"0.450 g","location":"specification"},{"text":"95°C","location":"specification"},{"text":"9.80 × 10^4 Pa","location":"specification"},{"text":"136 cm^3","location":"specification"}]},{"type":"text","order":1,"title":"Convert units to SI","content":"Convert temperature ($T$) to Kelvin and volume ($V$) to cubic meters ($m^3$) so they align with the SI units of the gas constant $R$:\n\n* **Temperature:**\n $$T = 95 + 273 = 368\\text{ K}$$\n\n* **Volume:**\n Since $1\\text{ m}^3 = 10^6\\text{ cm}^3$, we divide by $10^6$:\n $$V = 136 \\times 10^{-6}\\text{ m}^3 = 1.36 \\times 10^{-4}\\text{ m}^3$$\n\nThese conversions are essential to earn your first mark (**M1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$M = \\frac{\\htmlData{anchor=mass}{\\color{@sky}{m}}R\\htmlData{anchor=temp}{\\color{@amber}{T}}}{\\htmlData{anchor=pres}{\\color{@purple}{p}}\\htmlData{anchor=vol}{\\color{@green}{V}}}$"},{"latex":"$$M = \\frac{\\htmlData{anchor=mass-sub}{\\color{@sky}{0.450}} \\times 8.31 \\times \\htmlData{anchor=temp-sub}{\\color{@amber}{368}}}{\\htmlData{anchor=pres-sub}{\\color{@purple}{9.80 \\times 10^4}} \\times \\htmlData{anchor=vol-sub}{\\color{@green}{1.36 \\times 10^{-4}}}}$"},{"latex":"$$M \\approx 103.3\\text{ g mol}^{-1}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mass"},{"color":"amber","style":"text-color","anchor":"temp"},{"color":"purple","style":"text-color","anchor":"pres"},{"color":"green","style":"text-color","anchor":"vol"}],"connections":[{"to":"mass-sub","from":"mass","color":"sky","label":"mass in g","style":"solid"},{"to":"temp-sub","from":"temp","color":"amber","label":"temperature in K","style":"solid"},{"to":"pres-sub","from":"pres","color":"purple","label":"pressure in Pa","style":"solid"},{"to":"vol-sub","from":"vol","color":"green","label":"volume in m³","style":"solid"}]},"order":2,"title":"Calculate molar mass directly","description":"We rearrange the ideal gas equation $pV = nRT$ by substituting $n = \\frac{m}{M}$. This gives $pV = \\frac{mRT}{M}$, which rearranges directly to the molar mass formula $M = \\frac{mRT}{pV}$."},{"type":"text","order":3,"title":"Perform calculation using GDC","content":"To find the numerical value of $M$, evaluate the fraction carefully using your calculator. This step corresponds to earning **M2** and **M3**.","calculator":[{"gdc":{"expression":"(0.450 * 8.31 * 368) / (9.80e4 * 1.36e-4)","instructions":"Enter the calculation using the division operator, ensuring the standard notation is correct:"},"ti84":{"expression":"(0.450 * 8.31 * 368) / (9.80E4 * 1.36E-4)","instructions":"Enter the formula in your main screen using parentheses to keep the numerator and denominator separated:"},"desmos":{"expression":"\\frac{0.450 \\cdot 8.31 \\cdot 368}{9.80 \\cdot 10^4 \\cdot 1.36 \\cdot 10^{-4}}","instructions":"Type the division as a fraction to visually align the numerator and denominator:"},"description":"Calculate the direct molar mass of the compound"}]},{"type":"text","order":4,"title":"Determine the molecular formula","content":"We are given that compound $\\textbf{X}$ is a chloride of sulfur. Let's find how many chlorine atoms are present per sulfur atom:\n\n1. **Molar mass of sulfur (S):** $32.1\\text{ g mol}^{-1}$\n2. **Molar mass of chlorine (Cl):** $35.5\\text{ g mol}^{-1}$\n\nSubtract the mass of one sulfur atom from the total molar mass:\n\n$$103.3 - 32.1 = 71.2\\text{ g mol}^{-1}$$\n\nDivide the remaining mass by the molar mass of chlorine to find the number of chlorine atoms:\n\n$$\\text{Number of Cl atoms} = \\frac{71.2}{35.5} \\approx 2$$\n\nThus, there are 2 chlorine atoms for every 1 sulfur atom (**A1**).\n\n**Molecular Formula:**\n$$\\text{SCl}_2$$"}]},{"id":"4c69c009-0121-4f68-b9e9-f2e4bfb79a5b:backwards","label":"Hypothesis Testing","steps":[{"type":"text","order":0,"title":"Convert given conditions","content":"Before testing our hypotheses, we must convert the experimental conditions to SI units so they are compatible with the ideal gas equation, $pV = nRT$.\n\n* **Temperature ($T$):** \n $$T = 95^\\circ\\text{C} + 273 = 368\\text{ K}$$\n\n* **Pressure ($p$):** \n $$p = 9.80 \\times 10^4\\text{ Pa}$$","highlights":[{"text":"95^\\circ\\text{C}","location":"specification"},{"text":"9.80 \\times 10^4\\text{ Pa}","location":"specification"}]},{"type":"text","order":1,"title":"Define hypotheses for X","content":"We will assume potential molecular formulas for the sulfur chloride compound $\\textbf{X}$ and calculate their respective molar masses ($M$):\n\n* **Hypothesis 1:** $\\text{SCl}$\n $$M(\\text{SCl}) = 32.1 + 35.5 = 67.6\\text{ g mol}^{-1}$$\n\n* **Hypothesis 2:** $\\text{SCl}_2$\n $$M(\\text{SCl}_2) = 32.1 + 2(35.5) = 103.1\\text{ g mol}^{-1}$$\n\n* **Hypothesis 3:** $\\text{S}_2\\text{Cl}_2$\n $$M(\\text{S}_2\\text{Cl}_2) = 2(32.1) + 2(35.5) = 135.2\\text{ g mol}^{-1}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"For a sample of mass $m = 0.450\\text{ g}$, we calculate the number of moles ($n = \\frac{m}{M}$) for each candidate formula:"},{"latex":"$$$n(\\text{Hypothesis 1}) = \\frac{0.450\\text{ g}}{\\htmlData{anchor=M1}{\\color{@sky}{67.6\\text{ g mol}^{-1}}}} \\approx \\htmlData{anchor=n1}{\\color{@sky}{6.66 \\times 10^{-3}\\text{ mol}}}$$"},{"latex":"$$$n(\\text{Hypothesis 2}) = \\frac{0.450\\text{ g}}{\\htmlData{anchor=M2}{\\color{@amber}{103.1\\text{ g mol}^{-1}}}} \\approx \\htmlData{anchor=n2}{\\color{@amber}{4.36 \\times 10^{-3}\\text{ mol}}}$$"},{"latex":"$$$n(\\text{Hypothesis 3}) = \\frac{0.450\\text{ g}}{\\htmlData{anchor=M3}{\\color{@purple}{135.2\\text{ g mol}^{-1}}}} \\approx \\htmlData{anchor=n3}{\\color{@purple}{3.33 \\times 10^{-3}\\text{ mol}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"M1"},{"color":"sky","style":"text-color","anchor":"n1"},{"color":"amber","style":"text-color","anchor":"M2"},{"color":"amber","style":"text-color","anchor":"n2"},{"color":"purple","style":"text-color","anchor":"M3"},{"color":"purple","style":"text-color","anchor":"n3"}],"connections":[{"to":"n1","from":"M1","color":"sky","label":"n = m/M","style":"solid"},{"to":"n2","from":"M2","color":"amber","label":"n = m/M","style":"solid"},{"to":"n3","from":"M3","color":"purple","label":"n = m/M","style":"solid"}]},"order":2,"title":"Calculate theoretical moles","description":"Calculate the moles corresponding to $0.450\\text{ g}$ of each candidate sulfur chloride."},{"type":"text","order":3,"title":"Calculate theoretical volumes","content":"Using the ideal gas equation rearranged for volume ($V = \\frac{nRT}{p}$), where $R = 8.31\\text{ J K}^{-1}\\text{ mol}^{-1}$:\n\n* **For Hypothesis 1 ($\\text{SCl}$):**\n $$V = \\frac{6.66 \\times 10^{-3} \\times 8.31 \\times 368}{9.80 \\times 10^4} \\approx 2.08 \\times 10^{-4}\\text{ m}^3 = 208\\text{ cm}^3$$\n\n* **For Hypothesis 2 ($\\text{SCl}_2$):**\n $$V = \\frac{4.36 \\times 10^{-3} \\times 8.31 \\times 368}{9.80 \\times 10^4} \\approx 1.36 \\times 10^{-4}\\text{ m}^3 = 136\\text{ cm}^3$$\n\n* **For Hypothesis 3 ($\\text{S}_2\\text{Cl}_2$):**\n $$V = \\frac{3.33 \\times 10^{-3} \\times 8.31 \\times 368}{9.80 \\times 10^4} \\approx 1.04 \\times 10^{-4}\\text{ m}^3 = 104\\text{ cm}^3$$"},{"type":"text","order":4,"title":"Identify correct formula","content":"Comparing our computed volumes with the measured volume of $136\\text{ cm}^3$:\n\n* **Hypothesis 2** ($\\text{SCl}_2$) yields a theoretical volume of exactly $136\\text{ cm}^3$.\n\nTherefore, the molecular formula of compound $\\textbf{X}$ is **$\\text{SCl}_2$**.","highlights":[{"text":"136\\text{ cm}^3","location":"specification"}]}]}],"generatedAt":"2026-06-11T09:26:34.252Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"686a0dcf-531a-488c-a539-a62e271b323b","revisionId":"51f92194-a271-4ae0-bed7-50cebe48698e","questionNumber":null,"questionType":"LA","specification":"The label on a bottle of a preservative solution lists the mass of sulfur dioxide, \$\\text{SO}_2\$, in mg, in a typical serving of \$200\\text{ cm}^3\$.\n\nA student carries out an investigation to check the sulfur dioxide content in the solution.\n\nSulfur dioxide is oxidised by iodine according to the following equation:\n\\[\\text{SO}_2(\\text{aq}) + \\text{I}_2(\\text{aq}) + 2\\text{H}_2\\text{O}(\\text{l}) \\rightarrow \\text{SO}_4^{2-}(\\text{aq}) + 2\\text{I}^-(\\text{aq}) + 4\\text{H}^+(\\text{aq})\\]\n\nThe student dilutes a \$200\\text{ cm}^3\$ serving of the preservative solution with water to \$250.0\\text{ cm}^3\$ in a volumetric flask.\n\nThe student then titrates a \$25.0\\text{ cm}^3\$ volume of this diluted solution with a \$1.50 \\times 10^{-3}\\text{ mol dm}^{-3}\$ iodine solution, \$\\text{I}_2(\\text{aq})\$.\n\nThe mean titre of \$\\text{I}_2(\\text{aq})\$ is \$18.40\\text{ cm}^3\$.","difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"6d14b61c-d291-4b0e-80b1-9cc05cc724d6","label":null,"content":"Determine the mass, in mg, of sulfur dioxide in the \$200\\text{ cm}^3\$ serving of the preservative solution. Give your answer to 3 significant figures.","metadata":null,"markscheme":"$93","marks":6,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"step_by_step_stoichiometry","label":"Step-by-Step Stoichiometric Calculation","steps":[{"type":"text","order":0,"title":"Calculate moles of iodine","content":"First, we determine the number of moles of iodine, $\\text{I}_2$, used in the titration. We are given the concentration of the iodine solution and the mean titre volume.\n\nUsing the formula:\n$$\\text{moles} = \\text{concentration (mol dm}^{-3}\\text{)} \\times \\text{volume (dm}^3\\text{)}$$\n\nConvert the volume from $\\text{cm}^3$ to $\\text{dm}^3$ by dividing by $1000$:\n$$\\text{Volume of } \\text{I}_2 = \\frac{18.40}{1000} = 0.01840 \\text{ dm}^3$$\n\nNow, calculate the moles of $\\text{I}_2$:\n$$\\begin{aligned} n(\\text{I}_2) &= 1.50 \\times 10^{-3} \\text{ mol dm}^{-3} \\times 0.01840 \\text{ dm}^3 \\\\ &= 2.76 \\times 10^{-5} \\text{ mol} \\end{aligned}$$"},{"type":"text","order":1,"title":"Determine moles of SO₂","content":"Next, we look at the balanced chemical equation to find the stoichiometric ratio between sulfur dioxide and iodine:\n$$\\text{SO}_2(\\text{aq}) + \\text{I}_2(\\text{aq}) + 2\\text{H}_2\\text{O}(\text{l}) \\rightarrow \\text{SO}_4^{2-}(\\text{aq}) + 2\\text{I}^-(\\text{aq}) + 4\\text{H}^+(\\text{aq})$$\n\nThe reacting ratio of $\\text{SO}_2$ to $\\text{I}_2$ is $1:1$.\n\nTherefore, the moles of $\\text{SO}_2$ in the titrated $25.0\\text{ cm}^3$ aliquot is equal to the moles of $\\text{I}_2$:\n$$n(\\text{SO}_2)_{\\text{aliquot}} = 2.76 \\times 10^{-5} \\text{ mol}$$"},{"type":"text","order":2,"title":"Scale to total volume","content":"The $25.0\\text{ cm}^3$ aliquot titration sample was taken from a total diluted volume of $250.0\\text{ cm}^3$ in the volumetric flask.\n\nTo find the total amount of $\\text{SO}_2$ in the entire $250.0\\text{ cm}^3$ solution (which contains the original $200\\text{ cm}^3$ serving), we scale up by the dilution factor:\n$$\\begin{aligned} n(\\text{SO}_2)_{\\text{total}} &= n(\\text{SO}_2)_{\\text{aliquot}} \\times \\frac{250.0\\text{ cm}^3}{25.0\\text{ cm}^3} \\\\ &= 2.76 \\times 10^{-5} \\text{ mol} \\times 10 \\\\ &= 2.76 \\times 10^{-4} \\text{ mol} \\end{aligned}$$"},{"type":"text","order":3,"title":"Calculate molar mass of SO₂","content":"To convert the moles of sulfur dioxide to mass, we need its molar mass, $M(\\text{SO}_2)$. Using the relative atomic masses from the periodic table ($M_r(\\text{S}) = 32.1$ and $M_r(\\text{O}) = 16.0$):\n\n$$M(\\text{SO}_2) = 32.1 + (2 \\times 16.0) = 64.1 \\text{ g mol}^{-1}$$\n*(Note: Using $M_r(\\text{S}) = 32.0$ leading to $64.0 \\text{ g mol}^{-1}$ is also acceptable.)*"},{"type":"text","order":4,"title":"Calculate the mass","content":"Using the formula:\n$$\\text{mass (g)} = \\text{moles} \\times \\text{molar mass (g mol}^{-1}\\text{)}$$\n\nCalculate the mass in grams:\n$$\\begin{aligned} \\text{mass of } \\text{SO}_2 &= 2.76 \\times 10^{-4} \\text{ mol} \\times 64.1 \\text{ g mol}^{-1} \\\\ &= 0.017691 \\text{ g} \\end{aligned}$$"},{"type":"text","order":5,"title":"Convert to milligrams","content":"Finally, we convert the mass from grams to milligrams by multiplying by $1000$, and then round our final answer to 3 significant figures:\n$$\\begin{aligned} \\text{mass in mg} &= 0.017691 \\text{ g} \\times 1000 \\text{ mg g}^{-1} \\\\ &= 17.691 \\text{ mg} \\\\ &\\approx 17.7 \\text{ mg} \\end{aligned}$$\n\n*(If you used $M_r = 64.0 \\text{ g mol}^{-1}$, the final rounded value is $17.7 \\text{ mg}$ as well, since $0.017664 \\text{ g} \\times 1000 = 17.664 \\text{ mg} \\approx 17.7 \\text{ mg}$.)*"}]},{"id":"dimensional_analysis","label":"Dimensional Analysis","steps":[{"type":"text","order":0,"title":"Define the target units","content":"$94"},{"type":"text","order":1,"title":"Construct the calculation chain","content":"$95"},{"type":"text","order":2,"title":"Calculate the final mass","content":"Let's perform the continuous multiplication directly:\n\n$$\\text{Mass of } \\text{SO}_2 = 18.40 \\times \\frac{1.50 \\times 10^{-3}}{1000} \\times 1 \\times \\frac{250.0}{25.0} \\times 64.1 \\times 1000$$\n\nTo simplify:\n* The $\\frac{1}{1000}$ factor in the volume-to-litre conversion cancels directly with the $\\times 1000$ factor in the gram-to-milligram conversion.\n* The dilution factor is $\\frac{250.0}{25.0} = 10$.\n\nThis simplifies our expression to:\n\n$$\\text{Mass of } \\text{SO}_2 = 18.40 \\times (1.50 \\times 10^{-3}) \\times 10 \\times 64.1$$\n\n$$\\text{Mass of } \\text{SO}_2 = 1.76916 \\approx 17.7\\text{ mg } (\\text{to 3 sig. figs.})$$","calculator":[{"gdc":{"expression":"18.40 * 1.50 * 10^(-3) * 10 * 64.1","instructions":"Enter the following command on the main calculation screen:\n1. Type '18.40 * 1.50E-3 * 10 * 64.1'\n2. Press 'EXE'"},"ti84":{"expression":"18.40 * 1.50 * 10^(-3) * 10 * 64.1","instructions":"Enter the multiplication as written in the simplified form:\n1. Type '18.40 * 1.50 * 10^(-3) * 10 * 64.1'\n2. Press 'ENTER'"},"desmos":{"expression":"18.40 * (1.50 * 10^{-3}) * 10 * 64.1","instructions":"Enter the continuous multiplication chain in the input cell:"},"description":"Calculate the dimensional analysis multiplication directly."}]},{"type":"text","order":3,"title":"Explain markscheme allocation","content":"When sitting your exam, showing your steps clearly ensures you receive maximum marks even if you make an arithmetic error:\n\n* **M1**: Calculating the moles of iodine ($2.76 \\times 10^{-5}\\text{ mol}$)\n* **M2**: Deducing the $1:1$ stoichiometric ratio of iodine to sulfur dioxide\n* **M3**: Scaling up the aliquot to the total volumetric flask volume ($250.0 / 25.0 = 10 \\rightarrow 2.76 \\times 10^{-4}\\text{ mol}$)\n* **M4**: Correct calculation of the molar mass of $\\text{SO}_2$ ($64.1\\text{ g mol}^{-1}$)\n* **M5**: Calculating the mass of $\\text{SO}_2$ in grams ($0.0177\\text{ g}$)\n* **A1**: Expressing the final answer to **3 significant figures** with the correct unit ($17.7\\text{ mg}$)"}]}],"generatedAt":"2026-06-11T09:30:00.489Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"6d14b61c-d291-4b0e-80b1-9cc05cc724d6:step_by_step_stoichiometry","label":"Step-by-Step Stoichiometric Calculation","steps":[{"type":"text","order":0,"title":"Calculate moles of iodine","content":"First, we determine the number of moles of iodine, $\\text{I}_2$, used in the titration. We are given the concentration of the iodine solution and the mean titre volume.\n\nUsing the formula:\n$$\\text{moles} = \\text{concentration (mol dm}^{-3}\\text{)} \\times \\text{volume (dm}^3\\text{)}$$\n\nConvert the volume from $\\text{cm}^3$ to $\\text{dm}^3$ by dividing by $1000$:\n$$\\text{Volume of } \\text{I}_2 = \\frac{18.40}{1000} = 0.01840 \\text{ dm}^3$$\n\nNow, calculate the moles of $\\text{I}_2$:\n$$\\begin{aligned} n(\\text{I}_2) &= 1.50 \\times 10^{-3} \\text{ mol dm}^{-3} \\times 0.01840 \\text{ dm}^3 \\\\ &= 2.76 \\times 10^{-5} \\text{ mol} \\end{aligned}$$"},{"type":"text","order":1,"title":"Determine moles of SO₂","content":"Next, we look at the balanced chemical equation to find the stoichiometric ratio between sulfur dioxide and iodine:\n$$\\text{SO}_2(\\text{aq}) + \\text{I}_2(\\text{aq}) + 2\\text{H}_2\\text{O}(\text{l}) \\rightarrow \\text{SO}_4^{2-}(\\text{aq}) + 2\\text{I}^-(\\text{aq}) + 4\\text{H}^+(\\text{aq})$$\n\nThe reacting ratio of $\\text{SO}_2$ to $\\text{I}_2$ is $1:1$.\n\nTherefore, the moles of $\\text{SO}_2$ in the titrated $25.0\\text{ cm}^3$ aliquot is equal to the moles of $\\text{I}_2$:\n$$n(\\text{SO}_2)_{\\text{aliquot}} = 2.76 \\times 10^{-5} \\text{ mol}$$"},{"type":"text","order":2,"title":"Scale to total volume","content":"The $25.0\\text{ cm}^3$ aliquot titration sample was taken from a total diluted volume of $250.0\\text{ cm}^3$ in the volumetric flask.\n\nTo find the total amount of $\\text{SO}_2$ in the entire $250.0\\text{ cm}^3$ solution (which contains the original $200\\text{ cm}^3$ serving), we scale up by the dilution factor:\n$$\\begin{aligned} n(\\text{SO}_2)_{\\text{total}} &= n(\\text{SO}_2)_{\\text{aliquot}} \\times \\frac{250.0\\text{ cm}^3}{25.0\\text{ cm}^3} \\\\ &= 2.76 \\times 10^{-5} \\text{ mol} \\times 10 \\\\ &= 2.76 \\times 10^{-4} \\text{ mol} \\end{aligned}$$"},{"type":"text","order":3,"title":"Calculate molar mass of SO₂","content":"To convert the moles of sulfur dioxide to mass, we need its molar mass, $M(\\text{SO}_2)$. Using the relative atomic masses from the periodic table ($M_r(\\text{S}) = 32.1$ and $M_r(\\text{O}) = 16.0$):\n\n$$M(\\text{SO}_2) = 32.1 + (2 \\times 16.0) = 64.1 \\text{ g mol}^{-1}$$\n*(Note: Using $M_r(\\text{S}) = 32.0$ leading to $64.0 \\text{ g mol}^{-1}$ is also acceptable.)*"},{"type":"text","order":4,"title":"Calculate the mass","content":"Using the formula:\n$$\\text{mass (g)} = \\text{moles} \\times \\text{molar mass (g mol}^{-1}\\text{)}$$\n\nCalculate the mass in grams:\n$$\\begin{aligned} \\text{mass of } \\text{SO}_2 &= 2.76 \\times 10^{-4} \\text{ mol} \\times 64.1 \\text{ g mol}^{-1} \\\\ &= 0.017691 \\text{ g} \\end{aligned}$$"},{"type":"text","order":5,"title":"Convert to milligrams","content":"Finally, we convert the mass from grams to milligrams by multiplying by $1000$, and then round our final answer to 3 significant figures:\n$$\\begin{aligned} \\text{mass in mg} &= 0.017691 \\text{ g} \\times 1000 \\text{ mg g}^{-1} \\\\ &= 17.691 \\text{ mg} \\\\ &\\approx 17.7 \\text{ mg} \\end{aligned}$$\n\n*(If you used $M_r = 64.0 \\text{ g mol}^{-1}$, the final rounded value is $17.7 \\text{ mg}$ as well, since $0.017664 \\text{ g} \\times 1000 = 17.664 \\text{ mg} \\approx 17.7 \\text{ mg}$.)*"}]},{"id":"6d14b61c-d291-4b0e-80b1-9cc05cc724d6:dimensional_analysis","label":"Dimensional Analysis","steps":[{"type":"text","order":0,"title":"Define the target units","content":"$96"},{"type":"text","order":1,"title":"Construct the calculation chain","content":"$97"},{"type":"text","order":2,"title":"Calculate the final mass","content":"Let's perform the continuous multiplication directly:\n\n$$\\text{Mass of } \\text{SO}_2 = 18.40 \\times \\frac{1.50 \\times 10^{-3}}{1000} \\times 1 \\times \\frac{250.0}{25.0} \\times 64.1 \\times 1000$$\n\nTo simplify:\n* The $\\frac{1}{1000}$ factor in the volume-to-litre conversion cancels directly with the $\\times 1000$ factor in the gram-to-milligram conversion.\n* The dilution factor is $\\frac{250.0}{25.0} = 10$.\n\nThis simplifies our expression to:\n\n$$\\text{Mass of } \\text{SO}_2 = 18.40 \\times (1.50 \\times 10^{-3}) \\times 10 \\times 64.1$$\n\n$$\\text{Mass of } \\text{SO}_2 = 1.76916 \\approx 17.7\\text{ mg } (\\text{to 3 sig. figs.})$$","calculator":[{"gdc":{"expression":"18.40 * 1.50 * 10^(-3) * 10 * 64.1","instructions":"Enter the following command on the main calculation screen:\n1. Type '18.40 * 1.50E-3 * 10 * 64.1'\n2. Press 'EXE'"},"ti84":{"expression":"18.40 * 1.50 * 10^(-3) * 10 * 64.1","instructions":"Enter the multiplication as written in the simplified form:\n1. Type '18.40 * 1.50 * 10^(-3) * 10 * 64.1'\n2. Press 'ENTER'"},"desmos":{"expression":"18.40 * (1.50 * 10^{-3}) * 10 * 64.1","instructions":"Enter the continuous multiplication chain in the input cell:"},"description":"Calculate the dimensional analysis multiplication directly."}]},{"type":"text","order":3,"title":"Explain markscheme allocation","content":"When sitting your exam, showing your steps clearly ensures you receive maximum marks even if you make an arithmetic error:\n\n* **M1**: Calculating the moles of iodine ($2.76 \\times 10^{-5}\\text{ mol}$)\n* **M2**: Deducing the $1:1$ stoichiometric ratio of iodine to sulfur dioxide\n* **M3**: Scaling up the aliquot to the total volumetric flask volume ($250.0 / 25.0 = 10 \\rightarrow 2.76 \\times 10^{-4}\\text{ mol}$)\n* **M4**: Correct calculation of the molar mass of $\\text{SO}_2$ ($64.1\\text{ g mol}^{-1}$)\n* **M5**: Calculating the mass of $\\text{SO}_2$ in grams ($0.0177\\text{ g}$)\n* **A1**: Expressing the final answer to **3 significant figures** with the correct unit ($17.7\\text{ 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Atoms and reactions

Question 49

Atoms and reactions Questions

Atoms and reactions

Question 49

Atoms and reactions Questions