$1.22 \mathrm{g}$ of Benzoic acid is dissolved in $100 \mathrm{g}$ of Acetone and $100 \mathrm{g}$ of Benzene separately. Boiling point of the solution in Acetone increases by $0.17°\mathrm{C},$ while that of, in the Benzene increases by $0.13°\mathrm{C}; {\mathrm{K}}_{\mathrm{b}}$ for Acetone and Benzene are $1.7 \mathrm{K} \mathrm{kg}{ \mathrm{mol}}^{-1}$ and $2.6 \mathrm{K} \mathrm{kg}{ \mathrm{mol}}^{-1}$respectively. Molecular weight of Benzoic acid in Benzene is $\mathrm{X}$ and in Acetone is $\mathrm{Y}$. Calculate $\mathrm{X}+\mathrm{Y}$?

For a solution formed by mixing liquids $\mathrm{L}$ and $\mathrm{M},$ the vapour  pressure of $\mathrm{L}$ plotted against the mole  fraction of $\mathrm{M}$ in solution is shown in the following figure. Here ${\mathrm{x}}_{\mathrm{L}}$ and ${\mathrm{x}}_{\mathrm{M}}$ represent mole fractions of $\mathrm{L}$  and $\mathrm{M},$ respectively, in the solution. The correct statement(s) applicable to this system is (are):

The plot given below shows $\mathrm{P}-\mathrm{T}$ curves (where $\mathrm{P}$ is the pressure and $\mathrm{T}$ is the temperature) for two solvents $\mathrm{X}$ and $\mathrm{Y}$ and isomolal solutions of $\mathrm{NaCl}$ in these solvents. $\mathrm{NaCl}$ completely dissociates in both the solvents.

On addition of equal number of moles of a non-volatile solute $\mathrm{S}$ in equal amount (in $\mathrm{kg}$ ) of these solvents, the elevation of boiling point of solvent $\mathrm{X}$ is three times that of solvent $\mathrm{Y}.$ Solute $\mathrm{S}$ is known to undergo dimerization in these solvents. If the degree of dimerization is $0.7$ in solvent $\mathrm{Y},$ the degree of dimerization in solvent $\mathrm{X}$ is: