$A$ solution containing $12.5 \ g$ of non-electrolyte substance in $250 \ g$ of water shows boiling point elevation of $0.52 \ K$. Calculate the molar mass of the substance. $(K_b = 0.52 \ K \ kg \ mol^{-1})$ ......... $g \ mol^{-1}$.

The boiling point of water $(100\,^{\circ}C)$ becomes $100.52\,^{\circ}C$,if $3\,g$ of a non-volatile solute is dissolved in $200\,mL$ of water. The molecular weight of the solute is: (${K_b}$ for water is $0.6\,K\,kg\,mol^{-1}$)

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 molality of an aqueous dilute solution containing a non-volatile solute is $0.1 \ m$. What is the boiling temperature (in $^{\circ}C$) of the solution? (Boiling point elevation constant, $K_b = 0.52 \ K \ kg \ mol^{-1}$; boiling temperature of water $= 100^{\circ}C$).

$A$ solution of $1$ molal concentration of a solute will have maximum boiling point elevation when the solvent is

$A$ 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 $2.53 \ ^\circ C \ m^{-1}$.