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 $= 2.53 \ K \ kg \ mol^{-1}$)

$A$ solution containing $0.1050 \ g$ of a substance in $15.84 \ g$ of ether has a boiling point $1 \ ^\circ C$ higher than that of pure ether. What is the molar mass of the substance? [Given: Molar elevation constant for ether per $100 \ g = 21.6$]

When $3.3 \ g$ of an unknown substance is dissolved in $125 \ g$ of benzene (boiling point $80^\circ \text{C}$),the boiling point of the resulting solution is $80.66^\circ \text{C}$. What is the molar mass of the unknown substance? $(K_b = 3.28 \ \text{K kg mol}^{-1})$

Calculate the molality of a solution of a non-volatile solute if the boiling point elevation is $1.75 \ K$ and the $K_b$ for the solvent is $3 \ K \ kg \ mol^{-1}$. (in $m$)

Calculate the molality of a solution of a non-volatile solute if the boiling point of the solution and the molal elevation constant for the solvent are $319.8 \text{ K}$ and $2.5 \text{ K kg mol}^{-1}$ respectively. [Boiling point of pure solvent $= 319.5 \text{ K}$]

Calculate the molar mass of a non-volatile solute when $4 \text{ g}$ of it is dissolved in $100 \text{ g}$ of a solvent that boils at $319.4 \text{ K}$. Given: $K_b = 2.4 \text{ K kg mol}^{-1}$ and the boiling point of the pure solvent is $319 \text{ K}$.