The solution from the following with the highest depression in freezing point (lowest freezing point) is:

$PbCl_2$ is dissolved in water to make its saturated solution. What will be the freezing point of this solution?
Given: $K_f (H_2O) = 2 \ K \ kg \ mol^{-1}$,$K_{sp} (PbCl_2) = 4 \times 10^{-6}$
(Assume molarity to be equal to molality) $..... ^oC$

$1 \, \text{mole}$ of each of $A$ and $B$ form an ideal solution of vapour pressure $100 \, \text{mm Hg}$. Addition of $2 \, \text{moles}$ of $B$ to it decreases the vapour pressure by $20 \, \text{mm Hg}$. The vapour pressure of $A$ and $B$ in pure state are,respectively:

$1 \ g$ of silver gets distributed between $10 \ cm^3$ of molten zinc and $100 \ cm^3$ of molten lead at $800^{\circ} C$. The percentage of silver in the zinc layer is approximately

Assertion : Lowering of vapour pressure is directly proportional to osmotic pressure of the solution.
Reason : Osmotic pressure is a colligative property.

The elevation in boiling point for $1 \ m$ solution of non-volatile solute $A$ is $3 \ K$. The depression in freezing point for $2 \ m$ solution of $A$ in the same solvent is $6 \ K$. The ratio of $K_{b}$ and $K_{f}$ i.e.,$K_{b} / K_{f}$ is $1 : X$. The value of $X$ is [nearest integer].