The elevation in boiling point of a $0.25 \ molal$ aqueous solution of a substance is $(K_b = 0.52 \ K \ kg \ mol^{-1})$. (in $K$)

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. . . . . . .

$11.1 \ g$ of $CaCl_2$ is dissolved in $1 \ kg$ of water. Determine the elevation in boiling point of the solution. $[K_b = 0.5 \ K \ kg \ mol^{-1}]$ :-

If $K_b$ denotes the molal elevation constant of water,then the boiling point of an aqueous solution containing $36 \ g$ of glucose (molar mass $= 180 \ g \ mol^{-1}$) per $dm^3$ is:

$5 \ g$ of a non-volatile organic substance with molar mass $M_A$ is dissolved in $200 \ g$ of tetrahydrofuran. If $K_b$ is the molal elevation constant of tetrahydrofuran,then $\Delta T_b$ will be:

What is the elevation in boiling point for a one molal solution of a solute in a solvent called?