$A$ non-volatile, non-electrolyte solid solute when dissolved in $40 \text{ g}$ of a solvent, the vapour pressure of the solvent decreased from $760 \text{ mm Hg}$ to $750 \text{ mm Hg}$. If the same solution boils at $320 \text{ K}$, then the number of moles of the solvent present in the solution is . . . . . . . (Nearest integer) [Given: boiling point of the pure solvent = $319.5 \text{ K}$, $K_b$ of the solvent = $0.3 \text{ K kg mol}^{-1}$]

Pick the correct statement.

Calculate the osmotic pressure in $atm$ at $0\,^oC$ of a $0.18\, m$ aqueous solution of $KCl$ which has a freezing point of $-0.68\,^oC$. Assume the volume of the solution is equal to the volume of pure water. $(K_f = 1.86\,^oC\, m^{-1})$

Which of the following statements are not correct?
$A$. For water, magnitude of $K_b$ is more than the magnitude of $K_f$.
$B$. The elevation in boiling point of water when a non-volatile solute is added to it is larger in magnitude than its depression in freezing point.
$C$. Osmotic pressure measurement is preferred over any other colligative property to determine molar mass of proteins and polymers.
$D$. The dimerised form of benzoic acid in benzene is $C_6H_5 - C(=O) - OH \cdots O = C(OH) - C_6H_5$. Choose the correct answer from the options given below:

$A$ solution has a $1:4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20 \ ^oC$ are $440 \ mm \ Hg$ for pentane and $120 \ mm \ Hg$ for hexane. The mole fraction of pentane in the vapour phase would be

The ionization constant of a monobasic acid $HA$ is to be determined. If a $0.025 \ m$ aqueous solution of the acid freezes at $-0.060 \ ^{\circ}C$,calculate the ionization constant $K_a$. (Assume molality = molarity and $K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$).