The boiling point of $0.2 \, mol \, kg^{-1}$ solution of $X$ in water is greater than equimolal solution of $Y$ in water. Which one of the following statements is true in this case?

Two different solutions were prepared by dissolving sucrose in water having concentrations of $1 \ m$ and $2 \ m$ respectively. The value of $(T_2 - T_1)$ will be ........... $K$. (Given: $K_b$ for water $= 0.52 \ K \ kg \ mol^{-1}$)

The molal elevation constant is the ratio of the elevation in $B.P.$ to

The plot given below shows $P-T$ curves (where $P$ is the pressure and $T$ is the temperature) for two solvents $X$ and $Y$ and isomolal solutions of $NaCl$ in these solvents. $NaCl$ completely dissociates in both the solvents.
On addition of equal number of moles of a non-volatile solute $S$ in equal amount (in $kg$) of these solvents,the elevation of boiling point of solvent $X$ is three times that of solvent $Y$. Solute $S$ is known to undergo dimerization in these solvents. If the degree of dimerization is $0.7$ in solvent $Y$,the degree of dimerization in solvent $X$ is. . . . . . .

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

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 $2.53 \, K \, kg \, mol^{-1}$.