Calculate the molality of a solution of a nonvolatile solute having a boiling point elevation of $1.89 \ K$,if the boiling point elevation constant of the solvent is $3.15 \ K \ kg \ mol^{-1}$. (in $m$)

Vessel-$1$ contains $w_2 \ g$ of a non-volatile solute $X$ dissolved in $w_1 \ g$ of water. Vessel-$2$ contains $w_2 \ g$ of another non-volatile solute $Y$ dissolved in $w_1 \ g$ of water. Both the vessels are at the same temperature and pressure. The molar mass of $X$ is $80 \%$ of that of $Y$. The van't Hoff factor for $X$ is $1.2$ times that of $Y$ for their respective concentrations. The elevation of boiling point for the solution in Vessel-$1$ is . . . . . . $\%$ of the solution in Vessel-$2$.

$1 \ g$ of non-volatile non-electrolyte solute is dissolved in $100 \ g$ of two different solvents $A$ and $B$ whose ebullioscopic constants are in the ratio of $1 : 5$. The ratio of the elevation in their boiling points,$\frac{\Delta T_b (A)}{\Delta T_b (B)}$ is

Molal elevation constant is the elevation in boiling point produced by

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 $0.01 \ K$ $\left[K_{b}=0.50 \ K \ kg \ mol^{-1}\right]$.

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 $= 118^{\circ} C]$.