If 0 . 2 moles of sulphuric acid is poured in to 250 ml of water, calculate the concentration of the solution?

We know that concentration is dependent on the number of moles of solute and the volume of solution in $\left(\mathrm{L}\right)$ and it is given by:

Molarity$=\frac{\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{L}}$

Here, $\mathrm{n}=0.2$, $\mathrm{V}=250 \mathrm{ml}$

Therefore,

Molarity$=\frac{0.2}{1}\times 4=0.8 \mathrm{M}$

Hence, concentration of $250 \mathrm{ml}$ of water containing $0.2$ moles of sulphuric acid will be $0.8 \mathrm{M}$