$25 g$ of a solute of molar mass $250\mathrm{g}{\mathrm{mol}}^{-1}$ is dissolved in $100 ml$ of water to obtain a solution whose density is $1.25 g m{l}^{-1}$. The Molarity and molality of the solution are respectively.

No. of moles of solute $=\frac{25}{250}=0.1\text{ moles }$

Molarity $=\frac{\text{ No. of moles of solute }}{\text{ Volume of solvent in litre }}=\frac{0.1}{\frac{100}{1000}}=1\mathrm{M}$

molality $=\frac{\text{ no. of moles of solute }}{\text{ mass of solvent in }\mathrm{kg}}$

Mass $=d\times V$

$=1.25\times 100=125\mathrm{g}$

molality $=\frac{0.1}{125/1000}=0.8m$