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 82 – 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/q82\",\"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":81,"initialQuestionWindow":[{"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 ($17.7\\text{ mg}$)"}]}],"generatedAt":"2026-06-11T09:30:00.562Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"6a5ac883-ca0a-43eb-97ed-c5861a8fecf1","revisionId":"44c72ee0-4ccb-4877-a81d-bff386c07bcb","questionNumber":null,"questionType":"LA","specification":"Compound \$\\text{Y}\$ contains phosphorus and fluorine only, and has a boiling point of \$-101.8\\ ^\\circ\\text{C}\$. At room temperature, a \$0.660\\text{ g}\$ sample of gaseous compound \$\\text{Y}\$ is found to occupy a volume of \$0.180\\text{ dm}^3\$. Under these conditions, \$1\\text{ mol}\$ of gas molecules has a volume of \$24.0\\text{ dm}^3\$.","difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"c82e77cb-8754-4de8-8841-13b614fe4f0f","label":null,"content":"Determine the molecular formula of compound \$\\text{Y}\$.","metadata":null,"markscheme":"- Calculates the amount, in moles, of gas molecules: $n = \\frac{0.180}{24.0} = 0.00750\\text{ mol}$ M1\n- Calculates the molar mass ($M_r$) of compound $\\text{Y}$: $M_r = \\frac{0.660}{0.00750} = 88.0\\text{ g mol}^{-1}$ M1\n- Subtracts the atomic mass of phosphorus ($31.0$) and divides by the atomic mass of fluorine ($19.0$) to find the ratio of fluorine atoms: $\\frac{88.0 - 31.0}{19.0} = 3$ M1\n- Deduces that the molecular formula of compound $\\text{Y}$ is $\\text{PF}_3$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"moles_and_molar_mass","label":"Moles and Molar Mass","steps":[{"type":"text","order":0,"title":"Calculate the amount of gas","content":"First, we find the number of moles ($n$) of the gaseous compound $\\text{Y}$. We are given its volume at room temperature and the molar volume of a gas under these conditions:\n\n* Volume of gas ($V$) = $0.180\\text{ dm}^3$\n* Molar volume ($V_m$) = $24.0\\text{ dm}^3\\text{ mol}^{-1}$\n\nUsing the formula:\n\n$$n = \\frac{V}{V_m}$$\n\nSubstituting the values:\n\n$$n = \\frac{0.180\\text{ dm}^3}{24.0\\text{ dm}^3\\text{ mol}^{-1}} = 0.00750\\text{ mol}$$\n\nThis calculation earns you the first mark (\\textbf{M1}).","highlights":[{"text":"occupy a volume of 0.180 dm3","location":"specification"},{"text":"1 mol of gas molecules has a volume of 24.0 dm3","location":"specification"}]},{"type":"text","order":1,"title":"Determine the molar mass","content":"Next, we calculate the molar mass ($M$) of compound $\\text{Y}$ using the relationship between mass ($m$), moles ($n$), and molar mass:\n\n$$M = \\frac{m}{n}$$\n\nUsing the mass of the sample ($0.660\\text{ g}$) and the moles calculated in the previous step:\n\n$$M = \\frac{0.660\\text{ g}}{0.00750\\text{ mol}} = 88.0\\text{ g mol}^{-1}$$\n\nCalculating the correct molar mass earns the second mark (\\textbf{M1}).","highlights":[{"text":"0.660 g sample","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Compound Y contains only phosphorus (atomic mass = $31.0\\text{ g mol}^{-1}$) and fluorine (atomic mass = $19.0\\text{ g mol}^{-1}$). Let's represent the molecular formula as $\\text{P}_y\\text{F}_x$. Since phosphorus typically forms a single central atom in binary halides of this mass scale, we assume 1 atom of phosphorus ($y=1$):"},{"latex":"$$\\text{Molar mass of F atoms} = \\htmlData{anchor=tot-mass}{\\color{@sky}{88.0}} - \\htmlData{anchor=p-mass}{\\color{@amber}{31.0}} = \\htmlData{anchor=f-mass-tot}{\\color{@purple}{57.0}}\\text{ g mol}^{-1}$"},{"latex":"$$\\text{Number of F atoms } (x) = \\frac{\\htmlData{anchor=f-mass-tot-sub}{\\color{@purple}{57.0}}}{\\htmlData{anchor=f-mass-one}{\\color{@teal}{19.0}}} = \\htmlData{anchor=ratio-res}{\\color{@green}{3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"tot-mass"},{"color":"amber","style":"text-color","anchor":"p-mass"},{"color":"purple","style":"text-color","anchor":"f-mass-tot"},{"color":"teal","style":"text-color","anchor":"f-mass-one"},{"color":"green","style":"text-color","anchor":"ratio-res"}],"connections":[{"to":"f-mass-tot-sub","from":"f-mass-tot","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Deduce the fluorine ratio","description":"Subtracting the contribution of phosphorus from the total molar mass gives the total mass contributed by fluorine, which we then divide by fluorine's atomic mass to find the number of fluorine atoms."},{"type":"text","order":3,"title":"State the molecular formula","content":"Subtracting the molar mass of one phosphorus atom ($31.0\\text{ g mol}^{-1}$) from the total molar mass of the compound ($88.0\\text{ g mol}^{-1}$) leaves $57.0\\text{ g mol}^{-1}$ for the fluorine atoms (\\textbf{M1}).\n\nDividing this remaining mass by the atomic mass of fluorine ($19.0\\text{ g mol}^{-1}$):\n\n$$\\frac{57.0}{19.0} = 3$$\n\nTherefore, there are $3$ fluorine atoms for every $1$ phosphorus atom. The molecular formula of compound $\\text{Y}$ is:\n\n$$\\text{PF}_3$$\n\nStating the correct molecular formula awards the final mark (\\textbf{A1})."}]},{"id":"scaling_to_molar_volume","label":"Scaling to Molar Volume","steps":[{"type":"text","order":0,"title":"Identify the strategy","content":"We need to find the molecular formula of compound $\\text{Y}$, which contains only phosphorus ($\\text{P}$) and fluorine ($\\text{F}$).\n\nUsing the **Scaling to Molar Volume** approach, we know that under room conditions, exactly $1\\text{ mol}$ of any gas occupies a volume of $24.0\\text{ dm}^3$.\n\nTherefore, the mass of $24.0\\text{ dm}^3$ of compound $\\text{Y}$ is numerically equal to its molar mass ($M_r$) in $\\text{g mol}^{-1}$. We can calculate this using direct proportion.","highlights":[{"text":"Compound \$\\text{Y}\$ contains phosphorus and fluorine only","location":"specification"},{"text":"1\\text{ mol} of gas molecules has a volume of 24.0\\text{ dm}^3","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Volume of sample} = \\htmlData{anchor=vol-sample}{\\color{@sky}{0.180\\text{ dm}^3}}$"},{"latex":"$$\\text{Mass of sample} = \\htmlData{anchor=mass-sample}{\\color{@amber}{0.660\\text{ g}}}$"},{"latex":"$$\\text{Molar volume} = \\htmlData{anchor=vol-molar}{\\color{@purple}{24.0\\text{ dm}^3}}$"},{"latex":"$$M_r = \\frac{\\htmlData{anchor=mass-sub}{\\color{@amber}{0.660\\text{ g}}}}{\\htmlData{anchor=vol-sub}{\\color{@sky}{0.180\\text{ dm}^3}}} \\times \\htmlData{anchor=mvol-sub}{\\color{@purple}{24.0\\text{ dm}^3}}$"},{"latex":"$$M_r = \\htmlData{anchor=molar-mass}{\\color{@teal}{88.0\\text{ g mol}^{-1}}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"mass-sample"},{"color":"sky","style":"text-color","anchor":"vol-sample"},{"color":"purple","style":"text-color","anchor":"vol-molar"},{"color":"teal","style":"box","anchor":"molar-mass"}],"connections":[{"to":"mass-sub","from":"mass-sample","color":"amber","label":"mass of sample","style":"solid"},{"to":"vol-sub","from":"vol-sample","color":"sky","label":"sample volume","style":"solid"},{"to":"mvol-sub","from":"vol-molar","color":"purple","label":"molar volume","style":"solid"}]},"order":1,"title":"Scale to molar volume","description":"We scale the mass of our sample up to the volume occupied by one mole of gas ($24.0\\text{ dm}^3$) to find the molar mass directly."},{"type":"text","order":2,"title":"Deduce the formula","content":"Now we determine how many phosphorus ($A_r = 31.0$) and fluorine ($A_r = 19.0$) atoms make up our molar mass of $88.0\\text{ g mol}^{-1}$.\n\nLet the molecular formula be $\\text{P}_x\\text{F}_y$. \n\nSince phosphorus has a relatively large relative atomic mass, we can assume there is $1$ phosphorus atom ($x = 1$) because two phosphorus atoms would weigh $2 \\times 31.0 = 62.0\\text{ g mol}^{-1}$, leaving only $26.0\\text{ g mol}^{-1}$ for fluorine (which is not divisible by $19.0$).\n\nSubtract the mass of $1$ phosphorus atom from the molar mass:\n\n$$\\text{Mass due to fluorine} = 88.0 - 31.0 = 57.0\\text{ g mol}^{-1}$$\n\nNow, divide by the relative atomic mass of fluorine ($19.0$) to find the number of fluorine atoms ($y$):\n\n$$y = \\frac{57.0}{19.0} = 3$$\n\nThus, compound $\\text{Y}$ has $1$ phosphorus atom and $3$ fluorine atoms."},{"type":"text","order":3,"title":"State the final answer","content":"The molecular formula of compound $\\text{Y}$ is **$\\text{PF}_3$**.\n\n### Markscheme Breakdown\n- **M1**: Finding the ratio of volume / molar volume (equivalent to $0.00750\\text{ mol}$ of gas molecules).\n- **M1**: Calculating the molar mass ($88.0\\text{ g mol}^{-1}$).\n- **M1**: Subtracting the mass of phosphorus and dividing by the atomic mass of fluorine to find $3$ fluorine atoms.\n- **A1**: Correctly writing the molecular formula as $\\text{PF}_3$."}]},{"id":"testing_common_formulas","label":"Testing Common Fluorides","steps":[{"type":"text","order":0,"title":"Identify common phosphorus fluorides","content":"Since the gaseous compound $\\text{Y}$ contains only phosphorus and fluorine, we can list the most common stable phosphorus fluorides we might expect: \n* Phosphorus trifluoride ($\\text{PF}_3$)\n* Phosphorus pentafluoride ($\\text{PF}_5$)\n\nLet's calculate the molar masses ($M_r$) for both potential molecular formulas using the periodic table:\n\n* For $\\text{PF}_3$:\n $$M_r = 31.0 + 3 \\times 19.0 = 88.0\\text{ g mol}^{-1}$$\n* For $\\text{PF}_5$:\n $$M_r = 31.0 + 5 \\times 19.0 = 126.0\\text{ g mol}^{-1}$$"},{"type":"text","order":1,"title":"Calculate the moles of gas","content":"Using the given conditions, $1\\text{ mol}$ of gas occupies a volume of $24.0\\text{ dm}^3$ at room temperature and pressure. \n\nWe can find the number of moles ($n$) in our sample volume of $0.180\\text{ dm}^3$:\n\n$$n = \\frac{\\text{Volume}}{\\text{Molar volume}} = \\frac{0.180\\text{ dm}^3}{24.0\\text{ dm}^3\\text{ mol}^{-1}} = 0.00750\\text{ mol}$$","highlights":[{"text":"0.180 dm^3","location":"specification"},{"text":"1 mol of gas molecules has a volume of 24.0 dm^3","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$n = \\htmlData{anchor=mol-val}{\\color{@sky}{0.00750}} \\text{ mol}$"},{"latex":"$$\\text{Expected Mass } (m) = n \\times M_r$"},{"latex":"$$\\text{For } \\text{PF}_3: \\; m = \\color{@sky}{0.00750} \\times \\htmlData{anchor=pf3-mr}{\\color{@amber}{88.0}} = \\htmlData{anchor=pf3-mass}{\\color{@green}{0.660}} \\text{ g}$"},{"latex":"$$\\text{For } \\text{PF}_5: \\; m = \\color{@sky}{0.00750} \\times \\htmlData{anchor=pf5-mr}{\\color{@purple}{126.0}} = \\htmlData{anchor=pf5-mass}{\\color{@red}{0.945}} \\text{ g}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mol-val"},{"color":"amber","style":"text-color","anchor":"pf3-mr"},{"color":"green","style":"box","anchor":"pf3-mass"},{"color":"purple","style":"text-color","anchor":"pf5-mr"},{"color":"red","style":"text-color","anchor":"pf5-mass"}],"connections":[{"to":"pf3-mass","from":"mol-val","color":"sky","label":"multiply by Mr","style":"dashed"}]},"order":2,"title":"Calculate expected masses","description":"We now test our candidates to see which one matches the measured sample mass of $0.660\\text{ g}$."},{"type":"text","order":3,"title":"Match to given sample mass","content":"The calculated expected mass for a $0.180\\text{ dm}^3$ sample of $\\text{PF}_3$ is exactly $0.660\\text{ g}$, which matches our experimental data.\n\nTherefore, the molecular formula of compound $\\text{Y}$ is **$\\text{PF}_3$**.","highlights":[{"text":"0.660 g","location":"specification"}]}]}],"generatedAt":"2026-06-11T07:58:20.384Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c82e77cb-8754-4de8-8841-13b614fe4f0f:moles_and_molar_mass","label":"Moles and Molar Mass","steps":[{"type":"text","order":0,"title":"Calculate the amount of gas","content":"First, we find the number of moles ($n$) of the gaseous compound $\\text{Y}$. We are given its volume at room temperature and the molar volume of a gas under these conditions:\n\n* Volume of gas ($V$) = $0.180\\text{ dm}^3$\n* Molar volume ($V_m$) = $24.0\\text{ dm}^3\\text{ mol}^{-1}$\n\nUsing the formula:\n\n$$n = \\frac{V}{V_m}$$\n\nSubstituting the values:\n\n$$n = \\frac{0.180\\text{ dm}^3}{24.0\\text{ dm}^3\\text{ mol}^{-1}} = 0.00750\\text{ mol}$$\n\nThis calculation earns you the first mark (\\textbf{M1}).","highlights":[{"text":"occupy a volume of 0.180 dm3","location":"specification"},{"text":"1 mol of gas molecules has a volume of 24.0 dm3","location":"specification"}]},{"type":"text","order":1,"title":"Determine the molar mass","content":"Next, we calculate the molar mass ($M$) of compound $\\text{Y}$ using the relationship between mass ($m$), moles ($n$), and molar mass:\n\n$$M = \\frac{m}{n}$$\n\nUsing the mass of the sample ($0.660\\text{ g}$) and the moles calculated in the previous step:\n\n$$M = \\frac{0.660\\text{ g}}{0.00750\\text{ mol}} = 88.0\\text{ g mol}^{-1}$$\n\nCalculating the correct molar mass earns the second mark (\\textbf{M1}).","highlights":[{"text":"0.660 g sample","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Compound Y contains only phosphorus (atomic mass = $31.0\\text{ g mol}^{-1}$) and fluorine (atomic mass = $19.0\\text{ g mol}^{-1}$). Let's represent the molecular formula as $\\text{P}_y\\text{F}_x$. Since phosphorus typically forms a single central atom in binary halides of this mass scale, we assume 1 atom of phosphorus ($y=1$):"},{"latex":"$$\\text{Molar mass of F atoms} = \\htmlData{anchor=tot-mass}{\\color{@sky}{88.0}} - \\htmlData{anchor=p-mass}{\\color{@amber}{31.0}} = \\htmlData{anchor=f-mass-tot}{\\color{@purple}{57.0}}\\text{ g mol}^{-1}$"},{"latex":"$$\\text{Number of F atoms } (x) = \\frac{\\htmlData{anchor=f-mass-tot-sub}{\\color{@purple}{57.0}}}{\\htmlData{anchor=f-mass-one}{\\color{@teal}{19.0}}} = \\htmlData{anchor=ratio-res}{\\color{@green}{3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"tot-mass"},{"color":"amber","style":"text-color","anchor":"p-mass"},{"color":"purple","style":"text-color","anchor":"f-mass-tot"},{"color":"teal","style":"text-color","anchor":"f-mass-one"},{"color":"green","style":"text-color","anchor":"ratio-res"}],"connections":[{"to":"f-mass-tot-sub","from":"f-mass-tot","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Deduce the fluorine ratio","description":"Subtracting the contribution of phosphorus from the total molar mass gives the total mass contributed by fluorine, which we then divide by fluorine's atomic mass to find the number of fluorine atoms."},{"type":"text","order":3,"title":"State the molecular formula","content":"Subtracting the molar mass of one phosphorus atom ($31.0\\text{ g mol}^{-1}$) from the total molar mass of the compound ($88.0\\text{ g mol}^{-1}$) leaves $57.0\\text{ g mol}^{-1}$ for the fluorine atoms (\\textbf{M1}).\n\nDividing this remaining mass by the atomic mass of fluorine ($19.0\\text{ g mol}^{-1}$):\n\n$$\\frac{57.0}{19.0} = 3$$\n\nTherefore, there are $3$ fluorine atoms for every $1$ phosphorus atom. The molecular formula of compound $\\text{Y}$ is:\n\n$$\\text{PF}_3$$\n\nStating the correct molecular formula awards the final mark (\\textbf{A1})."}]},{"id":"c82e77cb-8754-4de8-8841-13b614fe4f0f:scaling_to_molar_volume","label":"Scaling to Molar Volume","steps":[{"type":"text","order":0,"title":"Identify the strategy","content":"We need to find the molecular formula of compound $\\text{Y}$, which contains only phosphorus ($\\text{P}$) and fluorine ($\\text{F}$).\n\nUsing the **Scaling to Molar Volume** approach, we know that under room conditions, exactly $1\\text{ mol}$ of any gas occupies a volume of $24.0\\text{ dm}^3$.\n\nTherefore, the mass of $24.0\\text{ dm}^3$ of compound $\\text{Y}$ is numerically equal to its molar mass ($M_r$) in $\\text{g mol}^{-1}$. We can calculate this using direct proportion.","highlights":[{"text":"Compound \$\\text{Y}\$ contains phosphorus and fluorine only","location":"specification"},{"text":"1\\text{ mol} of gas molecules has a volume of 24.0\\text{ dm}^3","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Volume of sample} = \\htmlData{anchor=vol-sample}{\\color{@sky}{0.180\\text{ dm}^3}}$"},{"latex":"$$\\text{Mass of sample} = \\htmlData{anchor=mass-sample}{\\color{@amber}{0.660\\text{ g}}}$"},{"latex":"$$\\text{Molar volume} = \\htmlData{anchor=vol-molar}{\\color{@purple}{24.0\\text{ dm}^3}}$"},{"latex":"$$M_r = \\frac{\\htmlData{anchor=mass-sub}{\\color{@amber}{0.660\\text{ g}}}}{\\htmlData{anchor=vol-sub}{\\color{@sky}{0.180\\text{ dm}^3}}} \\times \\htmlData{anchor=mvol-sub}{\\color{@purple}{24.0\\text{ dm}^3}}$"},{"latex":"$$M_r = \\htmlData{anchor=molar-mass}{\\color{@teal}{88.0\\text{ g mol}^{-1}}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"mass-sample"},{"color":"sky","style":"text-color","anchor":"vol-sample"},{"color":"purple","style":"text-color","anchor":"vol-molar"},{"color":"teal","style":"box","anchor":"molar-mass"}],"connections":[{"to":"mass-sub","from":"mass-sample","color":"amber","label":"mass of sample","style":"solid"},{"to":"vol-sub","from":"vol-sample","color":"sky","label":"sample volume","style":"solid"},{"to":"mvol-sub","from":"vol-molar","color":"purple","label":"molar volume","style":"solid"}]},"order":1,"title":"Scale to molar volume","description":"We scale the mass of our sample up to the volume occupied by one mole of gas ($24.0\\text{ dm}^3$) to find the molar mass directly."},{"type":"text","order":2,"title":"Deduce the formula","content":"Now we determine how many phosphorus ($A_r = 31.0$) and fluorine ($A_r = 19.0$) atoms make up our molar mass of $88.0\\text{ g mol}^{-1}$.\n\nLet the molecular formula be $\\text{P}_x\\text{F}_y$. \n\nSince phosphorus has a relatively large relative atomic mass, we can assume there is $1$ phosphorus atom ($x = 1$) because two phosphorus atoms would weigh $2 \\times 31.0 = 62.0\\text{ g mol}^{-1}$, leaving only $26.0\\text{ g mol}^{-1}$ for fluorine (which is not divisible by $19.0$).\n\nSubtract the mass of $1$ phosphorus atom from the molar mass:\n\n$$\\text{Mass due to fluorine} = 88.0 - 31.0 = 57.0\\text{ g mol}^{-1}$$\n\nNow, divide by the relative atomic mass of fluorine ($19.0$) to find the number of fluorine atoms ($y$):\n\n$$y = \\frac{57.0}{19.0} = 3$$\n\nThus, compound $\\text{Y}$ has $1$ phosphorus atom and $3$ fluorine atoms."},{"type":"text","order":3,"title":"State the final answer","content":"The molecular formula of compound $\\text{Y}$ is **$\\text{PF}_3$**.\n\n### Markscheme Breakdown\n- **M1**: Finding the ratio of volume / molar volume (equivalent to $0.00750\\text{ mol}$ of gas molecules).\n- **M1**: Calculating the molar mass ($88.0\\text{ g mol}^{-1}$).\n- **M1**: Subtracting the mass of phosphorus and dividing by the atomic mass of fluorine to find $3$ fluorine atoms.\n- **A1**: Correctly writing the molecular formula as $\\text{PF}_3$."}]},{"id":"c82e77cb-8754-4de8-8841-13b614fe4f0f:testing_common_formulas","label":"Testing Common Fluorides","steps":[{"type":"text","order":0,"title":"Identify common phosphorus fluorides","content":"Since the gaseous compound $\\text{Y}$ contains only phosphorus and fluorine, we can list the most common stable phosphorus fluorides we might expect: \n* Phosphorus trifluoride ($\\text{PF}_3$)\n* Phosphorus pentafluoride ($\\text{PF}_5$)\n\nLet's calculate the molar masses ($M_r$) for both potential molecular formulas using the periodic table:\n\n* For $\\text{PF}_3$:\n $$M_r = 31.0 + 3 \\times 19.0 = 88.0\\text{ g mol}^{-1}$$\n* For $\\text{PF}_5$:\n $$M_r = 31.0 + 5 \\times 19.0 = 126.0\\text{ g mol}^{-1}$$"},{"type":"text","order":1,"title":"Calculate the moles of gas","content":"Using the given conditions, $1\\text{ mol}$ of gas occupies a volume of $24.0\\text{ dm}^3$ at room temperature and pressure. \n\nWe can find the number of moles ($n$) in our sample volume of $0.180\\text{ dm}^3$:\n\n$$n = \\frac{\\text{Volume}}{\\text{Molar volume}} = \\frac{0.180\\text{ dm}^3}{24.0\\text{ dm}^3\\text{ mol}^{-1}} = 0.00750\\text{ mol}$$","highlights":[{"text":"0.180 dm^3","location":"specification"},{"text":"1 mol of gas molecules has a volume of 24.0 dm^3","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$n = \\htmlData{anchor=mol-val}{\\color{@sky}{0.00750}} \\text{ mol}$"},{"latex":"$$\\text{Expected Mass } (m) = n \\times M_r$"},{"latex":"$$\\text{For } \\text{PF}_3: \\; m = \\color{@sky}{0.00750} \\times \\htmlData{anchor=pf3-mr}{\\color{@amber}{88.0}} = \\htmlData{anchor=pf3-mass}{\\color{@green}{0.660}} \\text{ g}$"},{"latex":"$$\\text{For } \\text{PF}_5: \\; m = \\color{@sky}{0.00750} \\times \\htmlData{anchor=pf5-mr}{\\color{@purple}{126.0}} = \\htmlData{anchor=pf5-mass}{\\color{@red}{0.945}} \\text{ g}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mol-val"},{"color":"amber","style":"text-color","anchor":"pf3-mr"},{"color":"green","style":"box","anchor":"pf3-mass"},{"color":"purple","style":"text-color","anchor":"pf5-mr"},{"color":"red","style":"text-color","anchor":"pf5-mass"}],"connections":[{"to":"pf3-mass","from":"mol-val","color":"sky","label":"multiply by Mr","style":"dashed"}]},"order":2,"title":"Calculate expected masses","description":"We now test our candidates to see which one matches the measured sample mass of $0.660\\text{ g}$."},{"type":"text","order":3,"title":"Match to given sample mass","content":"The calculated expected mass for a $0.180\\text{ dm}^3$ sample of $\\text{PF}_3$ is exactly $0.660\\text{ g}$, which matches our experimental data.\n\nTherefore, the molecular formula of compound $\\text{Y}$ is **$\\text{PF}_3$**.","highlights":[{"text":"0.660 g","location":"specification"}]}]}],"generatedAt":"2026-06-11T07:58:20.435Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"6bce51a3-39bf-40a2-a374-88f8436c98d6","revisionId":"c03a9969-7a64-432a-b73b-6bde51d2b214","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"26eda376-328d-4aab-82a3-f63088908616","label":null,"content":"A student investigates the reaction between calcium carbonate and hydrochloric acid:\n\n\\[\\text{CaCO}_3(\\text{s}) + 2\\text{HCl}(\\text{aq}) \\rightarrow \\text{CaCl}_2(\\text{aq}) + \\text{H}_2\\text{O}(\\text{l}) + \\text{CO}_2(\\text{g})\\]\n\nThe student adds an excess of calcium carbonate to \$30.00\\text{ cm}^3\$ of \$\\text{HCl}(\\text{aq})\$ in a conical flask, which is quickly connected to a gas syringe.\n\nThe reaction is allowed to proceed until no more gas is evolved. The final volume of \$\\text{CO}_2\$ collected in the syringe is \$85\\text{ cm}^3\$, measured at room temperature and pressure (RTP).\n\nCalculate the initial concentration of the \$\\text{HCl}(\\text{aq})\$ solution. Give your answer to **two** significant figures. (Assume the molar gas volume at RTP is \$24.0\\text{ dm}^3\\text{ mol}^{-1}\$.)","metadata":null,"markscheme":"- Calculates the amount of $\\text{CO}_2$ gas collected: $n(\\text{CO}_2) = \\frac{85\\text{ cm}^3}{24000\\text{ cm}^3\\text{ mol}^{-1}} \\approx 3.54 \\times 10^{-3}\\text{ mol}$ M1\n- Deduces the stoichiometry of $2\\text{ mol}$ of $\\text{HCl}$ reacting to produce $1\\text{ mol}$ of $\\text{CO}_2$: $n(\\text{HCl}) = 2 \\times 3.54 \\times 10^{-3}\\text{ mol} \\approx 7.08 \\times 10^{-3}\\text{ mol}$ M1\n- Calculates the initial concentration of $\\text{HCl}$ using the solution volume of $30.00\\text{ cm}^3$ ($0.03000\\text{ dm}^3$): $[\\text{HCl}] = \\frac{7.08 \\times 10^{-3}\\text{ mol}}{0.03000\\text{ dm}^3} \\approx 0.236\\text{ mol dm}^{-3}$ M1\n- Rounds final answer to two significant figures to obtain $0.24\\text{ mol dm}^{-3}$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"stepwise_mole_calculation","label":"Step-by-step Mole Calculation","steps":[{"type":"text","order":0,"title":"Calculate moles of CO2","content":"First, calculate the number of moles of carbon dioxide gas, $\\text{CO}_2$, produced in the reaction. \n\nWe are given the volume of $\\text{CO}_2$ as $85\\text{ cm}^3$ and the molar gas volume at room temperature and pressure (RTP) as $24.0\\text{ dm}^3\\text{ mol}^{-1}$. To make our units compatible, convert the molar volume into $\\text{cm}^3\\text{ mol}^{-1}$:\n\n$$24.0\\text{ dm}^3\\text{ mol}^{-1} = 24.0 \\times 1000 = 24000\\text{ cm}^3\\text{ mol}^{-1}$$\n\nNow, calculate the moles ($n$) using the formula:\n\n$$n(\\text{CO}_2) = \\frac{\\text{Volume}}{\\text{Molar Gas Volume}} = \\frac{85\\text{ cm}^3}{24000\\text{ cm}^3\\text{ mol}^{-1}} \\approx 3.542 \\times 10^{-3}\\text{ mol}$$\n\n*(This step earns the first mark, **M1**)*","highlights":[{"text":"The final volume of \\text{CO}_2 collected in the syringe is 85\\text{ cm}^3","location":"specification"},{"text":"molar gas volume at RTP is 24.0\\text{ dm}^3\\text{ mol}^{-1}","location":"specification"}]},{"type":"text","order":1,"title":"Determine moles of HCl","content":"Next, use the stoichiometry of the balanced chemical equation to find how many moles of hydrochloric acid, $\\text{HCl}$, reacted:\n\n$$\\text{CaCO}_3(\\text{s}) + 2\\text{HCl}(\\text{aq}) \\rightarrow \\text{CaCl}_2(\\text{aq}) + \\text{H}_2\\text{O}(\\text{l}) + \\text{CO}_2(\\text{g})$$\n\nFrom the equation, the reacting ratio is:\n\n$$\\text{HCl} : \\text{CO}_2 = 2 : 1$$\n\nTherefore, the moles of $\\text{HCl}$ reacted is twice the moles of $\\text{CO}_2$ produced:\n\n$$n(\\text{HCl}) = 2 \\times 3.542 \\times 10^{-3}\\text{ mol} \\approx 7.083 \\times 10^{-3}\\text{ mol}$$\n\n*(This step earns the second mark, **M1**)*","highlights":[{"text":"\\text{CaCO}_3(\\text{s}) + 2\\text{HCl}(\\text{aq}) \\rightarrow \\text{CaCl}_2(\\text{aq}) + \\text{H}_2\\text{O}(\\text{l}) + \\text{CO}_2(\\text{g})","location":"specification"}]},{"type":"text","order":2,"title":"Calculate HCl concentration","content":"To find the initial concentration of the $\\text{HCl}(\\text{aq})$ solution, divide the moles of $\\text{HCl}$ by the volume of the solution in $\\text{dm}^3$.\n\nConvert the volume of $\\text{HCl}$ from $\\text{cm}^3$ to $\\text{dm}^3$:\n\n$$\\text{Volume} = \\frac{30.00\\text{ cm}^3}{1000} = 0.03000\\text{ dm}^3$$\n\nNow, calculate concentration ($c = \\frac{n}{V}$):\n\n$$[\\text{HCl}] = \\frac{7.083 \\times 10^{-3}\\text{ mol}}{0.03000\\text{ dm}^3} \\approx 0.2361\\text{ mol dm}^{-3}$$\n\n*(This calculation earns the third mark, **M1**)*","highlights":[{"text":"30.00\\text{ cm}^3","location":"specification"}]},{"type":"text","order":3,"title":"Round to significant figures","content":"Finally, the question asks us to give our answer to **two** significant figures:\n\n$$0.2361\\text{ mol dm}^{-3} \\approx 0.24\\text{ mol dm}^{-3}$$\n\nTherefore, the initial concentration of the hydrochloric acid solution is **$0.24\\text{ mol dm}^{-3}$**.\n\n*(This final rounded answer earns the accuracy mark, **A1**)*","highlights":[{"text":"two","location":"specification"}]}]},{"id":"dimensional_analysis","label":"Dimensional Analysis","steps":[{"type":"text","order":0,"title":"Analyze the goal","content":"We need to find the concentration of the $\\text{HCl}(\\text{aq})$ solution in units of $\\text{mol dm}^{-3}$ (or moles of $\\text{HCl}$ per $\\text{dm}^3$ of solution).\n\nTo do this in a single continuous calculation using **dimensional analysis**, we will start with our target quantity and chain together conversion factors that link the volume of gas produced to the volume of the acid reacted.","highlights":[{"text":"Calculate the initial concentration of the \\text{HCl}(\\text{aq}) solution. Give your answer to **two** significant figures.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Concentration of HCl} = \\frac{\\text{mol HCl}}{\\text{dm}^3 \\text{ HCl solution}}$"},{"latex":"$$= \\frac{\\htmlData{anchor=vol-co2}{\\color{@sky}{85\\text{ cm}^3\\text{ CO}_2}}}{\\htmlData{anchor=vol-hcl}{\\color{@pink}{30.00\\text{ cm}^3\\text{ HCl solution}}}} \\times \\frac{1\\text{ dm}^3\\text{ CO}_2}{\\htmlData{anchor=conv-vol}{\\color{@amber}{1000\\text{ cm}^3\\text{ CO}_2}}} \\times \\frac{1\\text{ mol CO}_2}{\\htmlData{anchor=molar-vol}{\\color{@purple}{24.0\\text{ dm}^3\\text{ CO}_2}}} \\times \\frac{\\htmlData{anchor=ratio-hcl}{\\color{@green}{2\\text{ mol HCl}}}}{\\htmlData{anchor=ratio-co2}{\\color{@green}{1\\text{ mol CO}_2}}} \\times \\frac{\\htmlData{anchor=conv-hcl}{\\color{@amber}{1000\\text{ cm}^3\\text{ HCl solution}}}}{1\\text{ dm}^3\\text{ HCl solution}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"vol-co2"},{"color":"pink","style":"box","anchor":"vol-hcl"},{"color":"amber","style":"underline","anchor":"conv-vol"},{"color":"purple","style":"underline","anchor":"molar-vol"},{"color":"green","style":"text-color","anchor":"ratio-hcl"},{"color":"green","style":"text-color","anchor":"ratio-co2"}],"connections":[{"to":"conv-vol","from":"vol-co2","color":"sky","label":"cancel cm³ CO₂","style":"dashed"},{"to":"conv-hcl","from":"vol-hcl","color":"pink","label":"cancel cm³ HCl","style":"dashed"}]},"order":1,"title":"Set up dimensional chain","description":"We establish a single mathematical chain. Notice how all units cancel out systematically, leaving only $\\text{mol HCl}$ in the numerator and $\\text{dm}^3\\text{ HCl solution}$ in the denominator."},{"type":"text","order":2,"title":"Simplify and compute","content":"Notice that the volume conversion factors ($1000\\text{ cm}^3 / 1\\text{ dm}^3$) in the numerator and denominator cancel each other out:\n\n$$\\text{Concentration of HCl} = \\frac{85}{30.00} \\times \\frac{1}{24.0} \\times \\frac{2}{1} \\text{ mol dm}^{-3}$$\n\nLet's evaluate this expression:\n\n$$\\text{Concentration} = \\frac{85 \\times 2}{30.00 \\times 24.0} = \\frac{170}{720} \\approx 0.23611\\text{ mol dm}^{-3}$$","calculator":[{"gdc":{"expression":"(85 * 2) / (30 * 24)","instructions":"Enter (85 * 2) / (30 * 24) on the main run screen."},"ti84":{"expression":"(85 * 2) / (30 * 24)","instructions":"Type (85 * 2) / (30 * 24) and press ENTER."},"desmos":{"expression":"\\frac{85 \\cdot 2}{30 \\cdot 24}","instructions":"Type the fraction expression into a blank line."},"description":"Calculate the final concentration fraction directly on your calculator."}]},{"type":"text","order":3,"title":"Round to significant figures","content":"The question specifies that we must round our final answer to **two** significant figures.\n\n$$0.23611\\text{ mol dm}^{-3} \\rightarrow 0.24\\text{ mol dm}^{-3}$$\n\n**Marks breakdown:**\n* **M1:** Determining the moles of $\\text{CO}_2$ (achieved via $\\frac{85}{24000} \\approx 3.54 \\times 10^{-3}\\text{ mol}$ equivalent within the chain).\n* **M1:** Applying the $1:2$ stoichiometric ratio to get moles of $\\text{HCl}$ ($7.08 \\times 10^{-3}\\text{ mol}$ equivalent within the chain).\n* **M1:** Dividing by the volume of acid ($0.03000\\text{ dm}^3$ equivalent within the chain).\n* **A1:** Correctly rounding the final answer to **0.24**.","highlights":[{"text":"Give your answer to **two** significant figures.","location":"specification"}]}]}],"generatedAt":"2026-06-11T08:06:35.877Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"26eda376-328d-4aab-82a3-f63088908616:stepwise_mole_calculation","label":"Step-by-step Mole Calculation","steps":[{"type":"text","order":0,"title":"Calculate moles of CO2","content":"First, calculate the number of moles of carbon dioxide gas, $\\text{CO}_2$, produced in the reaction. \n\nWe are given the volume of $\\text{CO}_2$ as $85\\text{ cm}^3$ and the molar gas volume at room temperature and pressure (RTP) as $24.0\\text{ dm}^3\\text{ mol}^{-1}$. To make our units compatible, convert the molar volume into $\\text{cm}^3\\text{ mol}^{-1}$:\n\n$$24.0\\text{ dm}^3\\text{ mol}^{-1} = 24.0 \\times 1000 = 24000\\text{ cm}^3\\text{ mol}^{-1}$$\n\nNow, calculate the moles ($n$) using the formula:\n\n$$n(\\text{CO}_2) = \\frac{\\text{Volume}}{\\text{Molar Gas Volume}} = \\frac{85\\text{ cm}^3}{24000\\text{ cm}^3\\text{ mol}^{-1}} \\approx 3.542 \\times 10^{-3}\\text{ mol}$$\n\n*(This step earns the first mark, **M1**)*","highlights":[{"text":"The final volume of \\text{CO}_2 collected in the syringe is 85\\text{ cm}^3","location":"specification"},{"text":"molar gas volume at RTP is 24.0\\text{ dm}^3\\text{ mol}^{-1}","location":"specification"}]},{"type":"text","order":1,"title":"Determine moles of HCl","content":"Next, use the stoichiometry of the balanced chemical equation to find how many moles of hydrochloric acid, $\\text{HCl}$, reacted:\n\n$$\\text{CaCO}_3(\\text{s}) + 2\\text{HCl}(\\text{aq}) \\rightarrow \\text{CaCl}_2(\\text{aq}) + \\text{H}_2\\text{O}(\\text{l}) + \\text{CO}_2(\\text{g})$$\n\nFrom the equation, the reacting ratio is:\n\n$$\\text{HCl} : \\text{CO}_2 = 2 : 1$$\n\nTherefore, the moles of $\\text{HCl}$ reacted is twice the moles of $\\text{CO}_2$ produced:\n\n$$n(\\text{HCl}) = 2 \\times 3.542 \\times 10^{-3}\\text{ mol} \\approx 7.083 \\times 10^{-3}\\text{ mol}$$\n\n*(This step earns the second mark, **M1**)*","highlights":[{"text":"\\text{CaCO}_3(\\text{s}) + 2\\text{HCl}(\\text{aq}) \\rightarrow \\text{CaCl}_2(\\text{aq}) + \\text{H}_2\\text{O}(\\text{l}) + \\text{CO}_2(\\text{g})","location":"specification"}]},{"type":"text","order":2,"title":"Calculate HCl concentration","content":"To find the initial concentration of the $\\text{HCl}(\\text{aq})$ solution, divide the moles of $\\text{HCl}$ by the volume of the solution in $\\text{dm}^3$.\n\nConvert the volume of $\\text{HCl}$ from $\\text{cm}^3$ to $\\text{dm}^3$:\n\n$$\\text{Volume} = \\frac{30.00\\text{ cm}^3}{1000} = 0.03000\\text{ dm}^3$$\n\nNow, calculate concentration ($c = \\frac{n}{V}$):\n\n$$[\\text{HCl}] = \\frac{7.083 \\times 10^{-3}\\text{ mol}}{0.03000\\text{ dm}^3} \\approx 0.2361\\text{ mol dm}^{-3}$$\n\n*(This calculation earns the third mark, **M1**)*","highlights":[{"text":"30.00\\text{ cm}^3","location":"specification"}]},{"type":"text","order":3,"title":"Round to significant figures","content":"Finally, the question asks us to give our answer to **two** significant figures:\n\n$$0.2361\\text{ mol dm}^{-3} \\approx 0.24\\text{ mol dm}^{-3}$$\n\nTherefore, the initial concentration of the hydrochloric acid solution is **$0.24\\text{ mol dm}^{-3}$**.\n\n*(This final rounded answer earns the accuracy mark, **A1**)*","highlights":[{"text":"two","location":"specification"}]}]},{"id":"26eda376-328d-4aab-82a3-f63088908616:dimensional_analysis","label":"Dimensional Analysis","steps":[{"type":"text","order":0,"title":"Analyze the goal","content":"We need to find the concentration of the $\\text{HCl}(\\text{aq})$ solution in units of $\\text{mol dm}^{-3}$ (or moles of $\\text{HCl}$ per $\\text{dm}^3$ of solution).\n\nTo do this in a single continuous calculation using **dimensional analysis**, we will start with our target quantity and chain together conversion factors that link the volume of gas produced to the volume of the acid reacted.","highlights":[{"text":"Calculate the initial concentration of the \\text{HCl}(\\text{aq}) solution. Give your answer to **two** significant figures.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Concentration of HCl} = \\frac{\\text{mol HCl}}{\\text{dm}^3 \\text{ HCl solution}}$"},{"latex":"$$= \\frac{\\htmlData{anchor=vol-co2}{\\color{@sky}{85\\text{ cm}^3\\text{ CO}_2}}}{\\htmlData{anchor=vol-hcl}{\\color{@pink}{30.00\\text{ cm}^3\\text{ HCl solution}}}} \\times \\frac{1\\text{ dm}^3\\text{ CO}_2}{\\htmlData{anchor=conv-vol}{\\color{@amber}{1000\\text{ cm}^3\\text{ CO}_2}}} \\times \\frac{1\\text{ mol CO}_2}{\\htmlData{anchor=molar-vol}{\\color{@purple}{24.0\\text{ dm}^3\\text{ CO}_2}}} \\times \\frac{\\htmlData{anchor=ratio-hcl}{\\color{@green}{2\\text{ mol HCl}}}}{\\htmlData{anchor=ratio-co2}{\\color{@green}{1\\text{ mol CO}_2}}} \\times \\frac{\\htmlData{anchor=conv-hcl}{\\color{@amber}{1000\\text{ cm}^3\\text{ HCl solution}}}}{1\\text{ dm}^3\\text{ HCl solution}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"vol-co2"},{"color":"pink","style":"box","anchor":"vol-hcl"},{"color":"amber","style":"underline","anchor":"conv-vol"},{"color":"purple","style":"underline","anchor":"molar-vol"},{"color":"green","style":"text-color","anchor":"ratio-hcl"},{"color":"green","style":"text-color","anchor":"ratio-co2"}],"connections":[{"to":"conv-vol","from":"vol-co2","color":"sky","label":"cancel cm³ CO₂","style":"dashed"},{"to":"conv-hcl","from":"vol-hcl","color":"pink","label":"cancel cm³ HCl","style":"dashed"}]},"order":1,"title":"Set up dimensional chain","description":"We establish a single mathematical chain. Notice how all units cancel out systematically, leaving only $\\text{mol HCl}$ in the numerator and $\\text{dm}^3\\text{ HCl solution}$ in the denominator."},{"type":"text","order":2,"title":"Simplify and compute","content":"Notice that the volume conversion factors ($1000\\text{ cm}^3 / 1\\text{ dm}^3$) in the numerator and denominator cancel each other out:\n\n$$\\text{Concentration of HCl} = \\frac{85}{30.00} \\times \\frac{1}{24.0} \\times \\frac{2}{1} \\text{ mol dm}^{-3}$$\n\nLet's evaluate this expression:\n\n$$\\text{Concentration} = \\frac{85 \\times 2}{30.00 \\times 24.0} = \\frac{170}{720} \\approx 0.23611\\text{ mol dm}^{-3}$$","calculator":[{"gdc":{"expression":"(85 * 2) / (30 * 24)","instructions":"Enter (85 * 2) / (30 * 24) on the main run screen."},"ti84":{"expression":"(85 * 2) / (30 * 24)","instructions":"Type (85 * 2) / (30 * 24) and press ENTER."},"desmos":{"expression":"\\frac{85 \\cdot 2}{30 \\cdot 24}","instructions":"Type the fraction expression into a blank line."},"description":"Calculate the final concentration fraction directly on your calculator."}]},{"type":"text","order":3,"title":"Round to significant figures","content":"The question specifies that we must round our final answer to **two** significant figures.\n\n$$0.23611\\text{ mol dm}^{-3} \\rightarrow 0.24\\text{ mol dm}^{-3}$$\n\n**Marks breakdown:**\n* **M1:** Determining the moles of $\\text{CO}_2$ (achieved via $\\frac{85}{24000} \\approx 3.54 \\times 10^{-3}\\text{ mol}$ equivalent within the chain).\n* **M1:** Applying the $1:2$ stoichiometric ratio to get moles of $\\text{HCl}$ ($7.08 \\times 10^{-3}\\text{ mol}$ equivalent within the chain).\n* **M1:** Dividing by the volume of acid ($0.03000\\text{ dm}^3$ equivalent within the chain).\n* **A1:** Correctly rounding the final answer to **0.24**.","highlights":[{"text":"Give your answer to **two** significant 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Atoms and reactions

Question 51

Atoms and reactions Questions

Atoms and reactions

Question 51

Atoms and reactions Questions