$2.5 \ g$ of a non-volatile,non-electrolyte is dissolved in $100 \ g$ of water at $25^{\circ} C$. The solution showed a boiling point elevation by $2^{\circ} C$. Assuming the solute concentration is negligible with respect to the solvent concentration,the vapour pressure of the resulting aqueous solution is . . . . . . $mm$ of $Hg$ (nearest integer).
[Given : Molal boiling point elevation constant of water $(K_b) = 0.52 \ K \ kg \ mol^{-1}$,
$1 \ atm$ pressure $= 760 \ mm$ of $Hg$,molar mass of water $= 18 \ g \ mol^{-1}]$

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:

$A$ 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 solubility product of a sparingly soluble $AB_2$ salt is $2.56 \times 10^{-4} \ M^3$ at $25^{\circ} C$. The $K_f$ of water is $1.8 \ K \ kg \ mol^{-1}$. The depression in freezing point of a saturated solution of $AB_2$ is (in $K$)

At $T(K)$,the vapour pressure of pure benzene (molar mass $= 78 \ g \ mol^{-1}$) is $0.85 \ bar$. When $2.0 \ g$ of a non-volatile,non-electrolyte solute is added to $39 \ g$ of benzene,the vapour pressure of the solution at $T(K)$ is $0.83 \ bar$. The elevation in boiling point (in $K$) of the same solution is: ($K_b$ of benzene is $2.6 \ K \ kg \ mol^{-1}$)

Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $300 \, K$ are $50.71 \, mm \, Hg$ and $32.06 \, mm \, Hg$ respectively. Calculate the mole fraction of benzene in the vapour phase if $80 \, g$ of benzene is mixed with $100 \, g$ of toluene.