$250\mathrm{ml}$ of a solution contains $6.3 \mathrm{g}$ of oxalic acid (MW=126). What is the volume (in liters) of water to be added to this solution to make it a 0.1 $\mathrm{N}$ solution

Formula of Normality $=\frac{\text{ number of equivalent of solute }}{1\text{ liter of solution }}$
$=\frac{\text{ Weight }}{\text{ eq.weight }}\times \frac{1000}{ \mathrm{V}\left(\mathrm{ml}\right)}$

By putting all the given values into the formula, we get-

$0·1=\frac{6·3}{\frac{126}{2}}\times \frac{1000}{ \mathrm{V}}$= 0.4

$N_1{V}_1=N_2{V}_2$

$V_2=1000ml$= 1 litre

The volume of water = Total Volume - Volume of oxalic acid 

= 1 - 0.25

= 0.75 liter