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.

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:

What is the boiling point of a decimolal aqueous solution of glucose if the molal elevation constant for water is $0.52 \ ^{\circ}C \ kg \ mol^{-1}$ (in $^{\circ}C$)?

If the boiling points of two solvents $X$ and $Y$ (having same molecular weights) are in the ratio $2 : 1$ and their enthalpies of vaporization are in the ratio $1 : 2$,then the boiling point elevation constant of $X$ is $\underline{ m }$ times the boiling point elevation constant of $Y$. The value of $m$ is $.....$. (Nearest integer)

$Y \ g$ of non-volatile organic substance of molecular mass $M$ is dissolved in $250 \ g$ benzene. The molal elevation constant of benzene is $K_b$. The elevation in its boiling point is given by:

If molality of a solution is $0.05 \ m$ and elevation in boiling point is $0.16 \ K$,then what is the molal elevation constant of the solvent?