Calculate the cryoscopic constant of solvent when 2.5 gram solute is dissolved in 35 gram solvent lowers its freezing point by 3 K. (molar mass of solute is $117 \mathrm{~g} \mathrm{~mol}^{-1}$)

Identify the reason for the solubility of polar solute in polar solvent from the following.

What is the osmotic pressure of solution prepared by dissolving 3 gram solute in $2 \mathrm{dm}^3$ water at 300 K . (Molar mass of solute $=60 \mathrm{~g} \mathrm{~mol}^{-1}$, $\mathrm{R}=0.0821 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K} \mathrm{~mol}^{-1})$

Calculate the molar mass of solute when 4 g of it dissolved in $1 \mathrm{dm}^3$ solvent has osmotic pressure 2 atm at 300 K . [R $\left[=0.082 \mathrm{~dm}^3 \mathrm{~atm} \mathrm{~K} \mathrm{~mol}^{-1}\right]$