Question

$49\%\left(w/w\right)H_{2}S{O}_4$ stock solution has a density of $1.2gm{l}^{-1}$. The volume of stock solution required to prepare 180 ml solution of 2N $H_{2}S{O}_4$ is:

• A. 20 ml
• B. 30 ml
• C. 12 ml
• D. 60 ml

Answer 1

The correct option is B 30 ml

49 gram $H_{2}S{O}_4\dots \dots$ 100 gram of solution
$\frac{49}{98}mole{H}_{2}S{O}_4\dots \dots \frac{100}{1.2}$ ml of solution
$\frac{1}{2}\times 1.2\times 10mole{H}_{2}S{O}_4\dots \dots 1000$ ml of solution
Molarity = 6M
Normality = n factor $\times$ molarity $=2\times 6=12N$
Now
$N_1{V}_1=N_2{V}_2$
$12N\times V_1=2N\times 180$
$V_1=\frac{2\times 180}{12}=30ml$