Question

$15\text{mL}$ of $\frac{\text{N}}{10}$ $NaOH$ solution completely neutralises $12\text{mL}$ of $H_{2}S{O}_4$​ solution. The normality of $H_{2}S{O}_4$​ solution will be

• A. $\frac{\text{N}}{8}$
• B. $\frac{\text{N}}{5}$
• C. $\text{N}$
• D. $\frac{\text{N}}{10}$

Answer 1

The correct option is A $\frac{\text{N}}{8}$

By law of equivalence:
$N_1{V}_1=N_2{V}_2$

$\frac{1}{10}\times 15=N_2\times 12$

$N_2=\frac{\text{N}}{8}$