Match the following.

List-$I$	List-$II$
$(A)$ Azeotrope	$(I)$ $\Delta T_b = i K_b m$
$(B)$ Henry's law	$(II)$ $p = K_H x$
$(C)$ Cryoscopic constant	$(III)$ $\Delta T_f / m$
$(D)$ Van't Hoff factor	$(IV)$ Deviation from Raoult's law
	$(V)$ $\pi = CRT$

The correct answer is

On dissolving $0.5 \ g$ of a non-volatile non-ionic solute to $39 \ g$ of benzene,its vapor pressure decreases from $650 \ mm \ Hg$ to $640 \ mm \ Hg$. The depression of freezing point of benzene (in $K$) upon addition of the solute is. . . . .
(Given data : Molar mass and the molal freezing point depression constant of benzene are $78 \ g \ mol^{-1}$ and $5.12 \ K \ kg \ mol^{-1}$,respectively)

$100 \ mL$ of $1.5\%(w/v)$ solution of urea has an osmotic pressure of $6.0 \ atm$ and $100 \ mL$ of $3.42\%(w/v)$ solution of cane sugar has an osmotic pressure of $2.4 \ atm$. If the two solutions are mixed,the osmotic pressure of the resulting solution in $atm$ is (Assume that there is no reaction between urea and cane sugar).

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 ..... $^o C$.

$1.22 \, g$ of an organic acid is separately dissolved in $100 \, g$ of benzene $(K_{b}=2.6 \, K \, kg \, mol^{-1})$ and $100 \, g$ of acetone $(K_{b}=1.7 \, K \, kg \, mol^{-1})$. The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by $0.17^{\circ} C$.
The increase in boiling point of solution in benzene in $^{\circ} C$ is $x \times 10^{-2}$. The value of $x$ is ..... .(Nearest integer)
$[$ Atomic mass : $C=12.0, H=1.0, O=16.0]$

$P_A = (235y - 125xy) \, \text{mm of Hg}$. $P_A$ is the partial pressure of $A$,$x$ is the mole fraction of $B$ in the liquid phase in the mixture of two liquids $A$ and $B$,and $y$ is the mole fraction of $A$ in the vapour phase. Then $P^o_B$ in $\text{mm of Hg}$ is: