At $T(\mathrm{~K}), 0.1$ mole of a non-volatile solute was dissolved in 0.9 mole of a volatile solvent. The vapour pressure of pure solvent is 0.9 bar. What is the vapour pressure (in bar ) of solution ?

At $300 \mathrm{~K}, 0.06 \mathrm{~kg}$ of an organic solute is dissolved in 1 kg water. The vapour pressure of solution at 300 K is 3.768 kPa . If vapour pressure of water at that temperature is 3.78 kPa , what is the molar mass of the organic solute (in $\mathrm{g} \mathrm{mol}^{-1}$ )? Assume the solution is dilute

Dry air contains $79 \% \mathrm{~N}_2$ and $21 \% \mathrm{O}_2$. At $T(\mathrm{~K})$, if Henry's law constants for $\mathrm{N}_2$ and $\mathrm{O}_2$ in water are $857 \times 10^4 \mathrm{~atm}$ and $4.56 \times 10^4 \mathrm{~atm}$, the ratio of mole fraction of $\mathrm{N}_2$ and $\mathrm{O}_2$ dissolved in water at 1 atm is