$22.7\text{mL}$ of a $\frac{N}{10}N{a}_{2}C{O}_3$ solution neutralizes $10.2\text{mL}$ of dilute $H_{2}S{O}_4$. Find the volume of water to be added to $400\text{mL}$ of the same $H_{2}S{O}_4$ solution to make its concentration exactly $\frac{N}{10}$.

The correct option is C $480\text{mL}$
We know that for neutralisation:
$(NV{)}_{N{a}_{2}C{O}_3}=(NV{)}_{H_{2}S{O}_4}$
$(0.1\times 22.7)=(N\times 10.2)$
Normality of $H_{2}S{O}_4=0.22\text{N}$
Now, this acid is diluted to a new concentration of $\frac{N}{10}$.
Let the amount of water to be added be $V_w$ mL
$N_{intial}{V}_{intial}=N_{final}{V}_{final}$

$(0.22\times 400)=(0.1\times (V_{water}+400\left)\right)$
On solving, $V_{water}=480\text{mL}$