Question

Normality of $10\left(W/V\right)H_{2}S{O}_4$ solution is nearly:

• A. $0.1$
• B. $0.2$
• C. $0.5$
• D. $2$

Answer 1

The correct option is D $2$

$10\%W/V=10g{H}_{2}S{O}_4$ in $100ml$ water

Normality $=\frac{weight\left(g\right)}{equivalentwt\left(g/eq\right)}\times \frac{1000}{Volumeofsolvent\left(mL\right)}$

$H_{2}S{O}_4$ is a dibasic acid ( 2 replaceable hydrogen's )

$\therefore$ equivalent weight $=\frac{molecularweight}{2}$

$=\frac{98.079}{2}$

$=49.0395g$

$\Rightarrow$ Normality $=\frac{10}{49.0395}\times \frac{100}{1000}=2.0369\simeq 2.$

Hence, the answer is $2.$