At $300 \ K$,an ideal solution is formed by mixing $460 \ g$ of toluene with $390 \ g$ of benzene. If the vapour pressure of pure toluene and pure benzene at $300 \ K$ are $32 \ mm$ and $40 \ mm$ respectively,the mole fraction of toluene in the vapour phase is:

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 $K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$).

An aqueous solution of urea (molar mass $= 60 \ g \ mol^{-1}$) boils at $100.18^o C$ at atmospheric pressure. If $K_f = 1.86 \ K \ kg \ mol^{-1}$ and $K_b = 0.512 \ K \ kg \ mol^{-1}$ for water,then the freezing point of the solution is equal to ..... $^o C$.

An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$

Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.
Reason : Raoult's law is a special case of Henry's law.

Given below are two statements.
Statement-$I$: Liquids $A$ and $B$ form a non-ideal solution with negative deviation. The interactions between $A$ and $B$ are weaker than $A-A$ and $B-B$ interactions.
Statement-$II$: In reverse osmosis,the applied pressure must be higher than the osmotic pressure of solution.
The correct answer is