At $40\,^oC$,the vapour pressure (in torr) of a solution of methyl alcohol $(A)$ and ethyl alcohol $(B)$ is represented by: $P_s = 120 X_A + 138$,where $X_A$ is the mole fraction of methyl alcohol. The values of $\lim_{X_A \to 0} (P_B^0)$ and $\lim_{X_B \to 0} (P_A^0)$ are:

After adding a non-volatile solute, the freezing point of water decreases to $-0.186^{\circ} C$. Calculate $\Delta T_b$ if $K_f = 1.86 \text{ K kg mol}^{-1}$ and $K_b = 0.521 \text{ K kg mol}^{-1}$. (in $\text{ K}$)

If a $1 \ m$ solution of benzoic acid in benzene has a freezing point depression of $2.56 \ ^{\circ}C$ $(K_f = 5.12 \ ^{\circ}C \ kg \ mol^{-1})$ and a boiling point elevation of $2.53 \ ^{\circ}C$ $(K_b = 2.53 \ ^{\circ}C \ kg \ mol^{-1})$,select the correct statement$(s)$:
Statement $I$: There is dimer formation when undergoing freezing.
Statement $II$: There is no change when undergoing boiling.
Statement $III$: Reverse of $I$ and $II$.
Statement $IV$: Dimer formation in freezing and boiling state.

At $T \ K$,the vapour pressure of pure benzene $(C_6H_6)$ and toluene $(C_7H_8)$ are $75 \ mm \ Hg$ and $22 \ mm \ Hg$ respectively. $23.4 \ g$ of benzene and $64.4 \ g$ of toluene are mixed to form an ideal solution. If the vapours are in equilibrium with the liquid mixture,the mole fraction of toluene in the vapour phase is (Atomic weight: $C=12, H=1$).

For a dilute solution containing $2.5 \ g$ of a non-volatile non-electrolyte solute in $100 \ g$ of water,the elevation in boiling point at $1 \ atm$ pressure is $2 \ ^\circ C$. Assuming the amount of solute is much lower than the amount of solvent,the vapour pressure $(mm \ Hg)$ of the solution is: (take $K_b = 0.76 \ K \ kg \ mol^{-1}$)

The $pH$ of a $0.1 \ M$ monobasic acid is $2$. Its osmotic pressure at a given temperature $T \ (K)$ is: (Given that the effective concentration for osmotic pressure is $(1+\alpha) \times$ concentration of acid, where $\alpha$ is the degree of dissociation)