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?

A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass $60 \mathrm{~g} \mathrm{~mol}^{-1}$ and 0.9 g of a non-volatile non-electrolyte solute ' B ' of molar mass $180 \mathrm{~g} \mathrm{~mol}^{-1}$ in $100 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}$ at $27^{\circ} \mathrm{C}$. Osmotic pressure of the solution will be

[Given: $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

' W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' $\mathrm{g} \mathrm{mol}^{-1}$ when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg . If aqueous solution of the electrolyte boils at 375 K and $\mathrm{K}_{\mathrm{b}}$ for water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, then the mole fraction of the electrolyte solute $\left(x_2\right)$ in the solution can be expressed as

(Given : density of water $=1 \mathrm{~g} / \mathrm{mL}$ and boiling point of water $=373 \mathrm{~K}$ )