When $3 \ g$ of a non-volatile solute is dissolved in $200 \ mL$ of water, the boiling point of the solution becomes $100.52 \ ^oC$. If $K_b$ for water is $0.6 \ K \ kg \ mol^{-1}$, the molar mass of the solute is ......... $g \ mol^{-1}$.

Boiling point of a $2\%$ aqueous solution of a non-volatile solute $A$ is equal to the boiling point of an $8\%$ aqueous solution of a non-volatile solute $B$. The relation between molecular weights of $A$ and $B$ is:

$A$ solution containing $2 \ g$ of a non-volatile solute in $20 \ g$ of water boils at $373.52 \ K$. The molecular mass of the solute is $....... \ g \ mol^{-1}$. (Nearest integer) Given,water boils at $373 \ K$,$K_b$ for water $= 0.52 \ K \ kg \ mol^{-1}$.

If the boiling point of a solution is $T_1$ and the boiling point of the pure solvent is $T_2$,then the elevation in boiling point is given by:

The plot given below shows $P-T$ curves (where $P$ is the pressure and $T$ is the temperature) for two solvents $X$ and $Y$ and isomolal solutions of $NaCl$ in these solvents. $NaCl$ completely dissociates in both the solvents.
On addition of equal number of moles of a non-volatile solute $S$ in equal amount (in $kg$) of these solvents,the elevation of boiling point of solvent $X$ is three times that of solvent $Y$. Solute $S$ is known to undergo dimerization in these solvents. If the degree of dimerization is $0.7$ in solvent $Y$,the degree of dimerization in solvent $X$ is. . . . . . .

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.