What is the boiling point of a solution of $0.1 \ m$ $KCl$ (in $K$)? $K_{b}$ of water is $0.52 \ K \ kg \ mol^{-1}$ and $\alpha = 100 \ \%$ (boiling point of pure water is $373 \ K$).

Calculate the molar mass of a nonvolatile solute if a solution containing $0.35 \ g$ of solute in $100 \ g$ of water has a boiling point elevation of $0.01 \ K$ $\left[K_{b}=0.50 \ K \ kg \ mol^{-1}\right]$.

$2 \, g$ 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)

$13.44 \ g$ of $CuCl_2$ is dissolved in $1 \ kg$ of water. Determine the elevation in boiling point of the solution. [$K_b = 0.52 \ K \ kg \ mol^{-1}$,$mol \ wt$ of $CuCl_2 = 134.4$]

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 $= 100^{\circ} C$)

Calculate the amount of water evaporated if $2.8 \ kg$ of $2 \ m$ $CaBr_2$ solution is heated to $110 \ ^oC$ at $1 \ atm$ .......... $gm$.
(Given: $K_b$ of water $= 0.5 \ K \ kg/mol$; Atomic mass of $Ca = 40$,$Br = 80$)