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Amount of substance

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Question 73

A student reacts 1.18 g of trimethylamine gas, (CH3)3N(\text{CH}_3)_3\text{N}(CH3​)3​N (Mr=59.0M_{\text{r}} = 59.0Mr​=59.0), with 40 cm3 of 0.80 mol dm-3 hydrochloric acid to produce a solution of trimethylammonium chloride.

(CH3)3N(g)+HCl(aq)→(CH3)3NH+(aq)+Cl−(aq) (\text{CH}_3)_3\text{N}(\text{g}) + \text{HCl}(\text{aq}) \rightarrow (\text{CH}_3)_3\text{NH}^+(\text{aq}) + \text{Cl}^-(\text{aq}) (CH3​)3​N(g)+HCl(aq)→(CH3​)3​NH+(aq)+Cl−(aq)

Assuming the volume of the solution does not change upon dissolving the gas, what is the concentration of trimethylammonium chloride in the resulting solution?

0.020.020.02

0.300.300.30

0.500.500.50

0.800.800.80

Amount of substance Questions

  1. A Level
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  3. /Amount of substance