Calculate the number of moles of nonvolatile solute dissolved in $0.5 \ kg$ solvent if molal elevation constant for solvent is $2 \ K \ kg \ mol^{-1}$ and the elevation in boiling point is $\Delta T_{b} = 0.8 \ K$.

The molal elevation constant is the ratio of elevation in boiling point to:

$1 \ g$ of non-volatile non-electrolyte solute is dissolved in $100 \ g$ of two different solvents $A$ and $B$ whose ebullioscopic constants are in the ratio of $1 : 5$. The ratio of the elevation in their boiling points,$\frac{\Delta T_b (A)}{\Delta T_b (B)}$ is

Calculate the molality of a solution of a non-volatile solute if the boiling point of the solution and the molal elevation constant for the solvent are $319.8 \text{ K}$ and $2.5 \text{ K kg mol}^{-1}$ respectively. [Boiling point of pure solvent $= 319.5 \text{ K}$]

When a substance is dissolved in a solvent,the vapour pressure of the solvent is decreased. This results in:

$A$ solution is prepared by dissolving a $2.5 \ g$ sample of an unknown compound in $34 \ g$ of benzene,$C_6H_6$. The solution boils at $1.38 \ ^\circ C$ higher than pure benzene. Which expression gives the molar mass of the unknown compound? $K_b$ of $C_6H_6$ is $2.53 \ ^\circ C \ m^{-1}$.