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, the amount of $NaOH$ in a $100$ ml mixture is:

• A. $0.2$ g
• B. $0.4$ g
• C. $0.6$ g
• D. $1.0$ g

Answer 1

Answer: (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 $0.1N$.
Thus, the mass of $NaOH$ $=0.1L\times 0.1N\times 40=0.4$ g.