The molal elevation constant is the ratio of elevation in boiling point to:

Sea water contains $29.25\%$ $NaCl$ and $19\%$ $MgCl_2$ by weight of solution. The normal boiling point of the sea water is $..........{}^{\circ}C$ (Nearest integer).
Assume $100\%$ ionization for both $NaCl$ and $MgCl_2$.
Given: $K_b(H_2O) = 0.52 \, K \, kg \, mol^{-1}$.
Molar mass of $NaCl$ and $MgCl_2$ is $58.5$ and $95 \, g \, mol^{-1}$ respectively.

The elevation in boiling point of a solution containing $1.8 \ g$ of glucose in $100 \ g$ of solvent is $0.1 \ ^\circ C$. Find the molal elevation constant of the liquid.

The correct relation between the elevation in boiling point and the molar mass of a solute is: ($W_1$ and $W_2$ are respectively weight of solvent $\&$ solute)

$0.1 \ molal$ aqueous solution of glucose boils at $100.16^{\circ} C$. What is the boiling point of a $0.5 \ molal$ aqueous solution of glucose (in $^{\circ} C$)?

If an aqueous solution contains $9 \%$ and $1 \%$ $(w/w)$ of two non-volatile non-electrolytes $X$ (molecular weight $180$) and $Y$ (molecular weight $50$) respectively,the boiling point of the solution in ${ }^{\circ} C$ is approximately $(K_b = 0.52 \ K \ kg \ mol^{-1})$