Question

The density of a stock solution of $49\%\left(W/W\right)H_{2}S{O}_4$ is $1.5g/ml$. The volume of this stock solution required to prepare $2.5L$ of $0.1M{H}_{2}S{O}_4$ solution is?

• A. $33.33ml$
• B. $25.55ml$
• C. $43.23ml$
• D. $38.25ml$

Answer 1

The correct option is A $33.33ml$

$1500g$ for $1L$ solution,

But it is $49\%$ $H_{2}S{O}_4$,

$\therefore$ weight of $H_{2}S{O}_4=0.49\times 1500=735g$

$\therefore$ moles of $H_{2}S{O}_4=\frac{735}{98}=7.5$

$\therefore$ Molarity $=\frac{7.5moles}{1L}=7.5M$

This is the molarity of stock solution,

$\because M_1{V}_1=M_2{V}_2$

$\therefore 7.5\times V=2.5\times 0.1$

$V=33.33mL$