Pure benzene boils at $80\,^oC$. When $1\,g$ of a solute is dissolved in $83.4\,g$ of benzene,the boiling point of the solution becomes $80.175\,^oC$. If the latent heat of vaporization of benzene is $90\,cal/g$,calculate the molar mass of the solute.

$11.1 \ g$ of $CaCl_2$ is dissolved in $1 \ kg$ of water. Determine the elevation in boiling point of the solution. $[K_b = 0.5 \ K \ kg \ mol^{-1}]$ :-

When $10 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of benzene,it raises the boiling point by $1 \ ^{\circ}C$. The molecular mass of the solute is ....... $g \ mol^{-1}$. ($K_b$ for benzene $= 2.53 \ K \ kg \ mol^{-1}$)

If the solution boils at a temperature $T_1$ and the solvent at a temperature $T_2$,the elevation of boiling point is given by:

The elevation in boiling point of a solution containing $1.8 \ g$ of glucose in $100 \ g$ of solvent is $0.1^{\circ}C$. The value of $K_b$ for the solvent is ........ $\frac{K}{m}$.

Vessel-$1$ contains $w_2 \ g$ of a non-volatile solute $X$ dissolved in $w_1 \ g$ of water. Vessel-$2$ contains $w_2 \ g$ of another non-volatile solute $Y$ dissolved in $w_1 \ g$ of water. Both the vessels are at the same temperature and pressure. The molar mass of $X$ is $80 \%$ of that of $Y$. The van't Hoff factor for $X$ is $1.2$ times that of $Y$ for their respective concentrations. The elevation of boiling point for the solution in Vessel-$1$ is . . . . . . $\%$ of the solution in Vessel-$2$.