Question

A sample of $H_{2}SO_{4}$ (density $1.787gmL^{- 1}$ ) is labelled as $86\%$ by weight. What volume $\left(\right.inmL\left.\right)$ of this acid has to be used to make $1L\text{of}0.2MH_{2}SO_{4}$ ?
[Report your answer to the nearest integer]

Answer 1

$H_{2}SO_{4}$ is 86% by weight
So weight of $H_{2}SO_{4}$ = 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 $H_{2}SO_{4}$ are used to prepare 1 L of 0.2 M $H_{2}SO_{4}$ .
So mmoles of concentrated $H_{2}SO_{4}=$ mmoles of dilute $H_{2}SO_{4}$
So $V\times 15.68=1000\times 0.2$
So V = 12.75 mL $\cong13mL$ .