An aqueous dilute solution containing non-volatile solute boils at $100.52^{\circ} C$. What is the molality of the solution (in $m$)? ($K_b = 0.52 \ K \ kg \ mol^{-1}$, boiling temperature of water $= 100^{\circ} C$)

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 ........ $\frac{K}{m}$.

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

The molal elevation boiling point constant for water is $0.513 \ ^{\circ}C \ kg \ mol^{-1}$. Calculate the boiling point of the solution if $0.1 \ mol$ of sugar is dissolved in $200 \ g$ of water. (in $^{\circ}C$)

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.

If the boiling point of water is $100 \,^{\circ}C$,how many grams of $NaCl$ must be added to $500 \,g$ of water to increase the boiling point by approximately $1 \,^{\circ}C$ (in $,g$)? Given $(K_{b})_{H_2O} = 0.52 \,K \cdot kg/mol$.