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 molality of the solution containing a nonvolatile solute if the boiling point elevation of the solution is $0.39 \ K$.
[$K_{b}$ of water $= 0.52 \ K \ kg \ mol^{-1}$]

The boiling point of benzene is $353.23 \, K$. When $1.80 \, g$ of a non-volatile solute was dissolved in $90 \, g$ of benzene,the boiling point is raised to $354.11 \, K$. Calculate the molar mass of the solute. $K_{b}$ for benzene is $2.53 \, K \, kg \, mol^{-1}$.

Elevation in boiling point was $0.52 \, ^\circ C$ when $6 \, g$ of a compound $X$ was dissolved in $100 \, g$ of water. The molecular weight of $X$ is: ($K_b$ for water is $0.52 \, K \, kg \, mol^{-1}$)

When a substance is dissolved in a solvent,the vapour pressure of the solvent decreases. It brings:

When $0.5 \ g$ of anthracene is dissolved in $35 \ g$ of chloroform,the boiling point increases by $0.3 \ K$. If the $K_b$ value for $CHCl_3$ is $3.9 \ K \ kg \ mol^{-1}$,the experimental molar mass of anthracene is ......... $g \ mol^{-1}$.