Consider the following aqueous solutions.

I. 2.2 g Glucose in 125 mL of solution.

II. 1.9 g Calcium chloride in 250 mL of solution.

III. 9.0 g Urea in 500 mL of solution.

IV. 20.5 g Aluminium sulphate in 750 mL of solution.

The correct increasing order of boiling point of these solutions will be :

[Given : Molar mass in g mol⁻¹ : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and S = 32]

At $\mathrm{T}(\mathrm{K}), 2$ moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg . At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg . The vapour pressure (in mm Hg ) of A and B are respectively:

At 298 K , the mole percentage of $\mathrm{N}_2(\mathrm{~g})$ in air is $80 \%$. Water is in equilibrium with air at a pressure of 10 atm . What is the mole fraction of $\mathrm{N}_2(\mathrm{~g})$ in water at 298 K ? $\left(\mathrm{K}_{\mathrm{H}}\right.$ for $\mathrm{N}_2$ is $\left.6.5 \times 10^7 \mathrm{~mm} \mathrm{Hg}\right)$

Two liquids A and B form an ideal solution at temperature T K . At T K , the vapour pressures of pure A and B are 55 and $15 \mathrm{kN} \mathrm{m}^{-2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?