$11.1 \ g$ of $CaCl_2$ is dissolved in $1 \ kg$ of water. Determine the elevation in boiling point of the solution. $[K_b = 0.5 \ K \ kg \ mol^{-1}]$ :-

Calculate the molar mass of a non-volatile solute when $5 \ g$ of it is dissolved in $50 \ g$ of solvent,which boils at $119.6^{\circ} C$. $[K_{b} = 3.2 \ K \ kg \ mol^{-1}$,boiling point of pure solvent $= 118^{\circ} C]$.

$2 \, g$ of a non-volatile non-electrolyte solute is dissolved in $200 \, g$ of two different solvents $A$ and $B$ whose ebullioscopic constants are in the ratio of $1: 8$. The elevation in boiling points of $A$ and $B$ are in the ratio $\frac{x}{y} (x: y)$. The value of $y$ is .... (Nearest integer)

If $0.15 \ g$ of a solute is dissolved in $15 \ g$ of solvent and the solution boils at a temperature $0.216 \ ^oC$ higher than that of the pure solvent, what is the molecular weight of the substance? (Given: molal elevation constant for the solvent is $2.16 \ ^oC \ kg \ mol^{-1}$)

If the enthalpy of vaporization is approximately the same,for which of the following will the value of the ebullioscopic constant $(K_b)$ be maximum?

The boiling point of a solution of $0.11 \ g$ of a substance in $15 \ g$ of ether was found to be $0.1 \ ^\circ C$ higher than that of the pure ether. The molecular weight of the substance will be $(K_b = 2.16 \ K \ kg \ mol^{-1})$.