Question

$0.106$ g of $N{a}_{2}C{O}_3$ completely neutralises $40.0$ mL of $H_{2}S{O}_4$. Hence, normality of $H_{2}S{O}_4$ solution is (write $0.06$ as $6$ in the answer) :

Answer 1

$0.106$ g $N{a}_{2}C{O}_3$ $=$ $\frac{0.106}{106}$ mol $=2\times {10}^{-3}$ equivalents
Equivalents of $H_{2}S{O}_4$ $=\frac{40N}{1000}$ equivalents, where $N$ is the normality.
Both have same number of equivalents in the reaction.
$\therefore \frac{40N}{1000}=$ $2\times {10}^{-3}$
$⟹N=0.05N$ $H_{2}S{O}_4$
Hence, the answer is $5$.