Question

$100$ mL of a mixture of $NaOH$ and $N{a}_{2}S{O}_4$ is neutralised by $10$ mL of $0.5M$ $H_{2}S{O}_4$. Hence, $NaOH$ in $100$ mL solution is:

• A. $0.2$ g
• B. $0.4$ g
• C. $0.6$ g
• D. none of the above

Answer 1

The correct option is B $0.4$ g

$N{a}_{2}S{O}_4$ does not take part in the reaction.

The correlation between the normality and volume of acid and base is $N_1{V}_1=N_2{V}_2$

Substituting values in the above expression, we get

$N_1\times 100=(2\times 0.5)\times 10$

Hence, the normality of NaOH solution is 0.1 N.

Thus, the mass of NaOH = 0.1L × 0.1N × 40 = 0.4 g.

Hence, option B is correct .