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:

$A$ solution of urea (molar mass $60 \ g \ mol^{-1}$) boils at $100.20^{\circ}C$ at atmospheric pressure. If $K_{f}$ and $K_{b}$ for water are $1.86$ and $0.512 \ K \ kg \ mol^{-1}$ respectively,the freezing point of the solution will be:

Evaluate the following statements for their correctness.
$(A)$ The elevation in boiling point temperature of water will be same for $0.1 \ M \ NaCl$ and $0.1 \ M$ urea.
$(B)$ Azeotropic mixtures boil without change in their composition.
$(C)$ Osmosis always takes place from hypertonic to hypotonic solution.
$(D)$ The density of $32 \% \ H_2SO_4$ solution having molarity $4.09 \ M$ is approximately $1.26 \ g \ mL^{-1}$.
$(E)$ $A$ negatively charged sol is obtained when $KI$ solution is added to silver nitrate solution.
Choose the correct answer from the options given below:

Elevation in the boiling point for $1 \ m$ solution of glucose is $2 \ K$. The depression in the freezing point for $2 \ m$ solution of glucose in the same solvent is $2 \ K$. The relation between $K_b$ and $K_f$ is

An aqueous solution freezes at $-0.186 \ ^\circ C$ ($K_f = 1.86 \ ^\circ C \ kg \ mol^{-1}$; $K_b = 0.512 \ ^\circ C \ kg \ mol^{-1}$). What is the elevation in boiling point of the same solution?

An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:
Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.