Question

A sample of $H_{2}S{O}_4$ (density $1.787gm{L}^{-1}$ ) is labelled as $86\%$ by weight. What volume $(inmL)$ of this acid has to be used to make $1L\text{of}0.2M{H}_{2}S{O}_4$ ?
[Report your answer to the nearest integer]

Answer 1

Answer: 13

$H_{2}S{O}_4$ is 86% by weight
So weight of $H_{2}S{O}_4$ = 86 g
Weight of solution = 100 g
So volume of solution $=\frac{100}{1.787}mL=\frac{100}{1.787\times 1000}L$
$M_{H_{2}S{O}_4}=\frac{86}{98\times \frac{100}{1.787\times 1000}}=15.68$
Let V mL of this $H_{2}S{O}_4$ are used to prepare 1 L of 0.2 M $H_{2}S{O}_4$ .
So mmoles of concentrated $H_{2}S{O}_4=$ mmoles of dilute $H_{2}S{O}_4$
So $V\times 15.68=1000\times 0.2$
So V = 12.75 mL $\cong 13mL$ .