Vapour pressure of solvent $A$ is $0.90 \text{ atm}$. When a non-volatile solute is added,the vapour pressure drops to $0.60 \text{ atm}$. What is the mole fraction of $A$ in the solution?

Dry air is passed through a solution containing $10 \ g$ of solute in $90 \ g$ of water and then through pure water. The loss in weight of the solution is $2.5 \ g$ and the loss in weight of the solvent is $0.05 \ g$. The molecular weight of the solute is:

Two liquids $A$ and $B$ have $P_A^o$ and $P_B^o$ in the ratio of $1 : 3$ and the ratio of number of moles of $A$ and $B$ in the liquid phase is $1 : 3$. The mole fraction of $A$ in the vapour phase in equilibrium with the solution is equal to:

Vapour pressure of a pure solvent at $90\,^{\circ}C$ is $1020\, torr$. $A$ solution of $0.1\, m$ concentration has a vapour pressure of $1000\, torr$. What is the molecular weight of the solvent?

In a solution,the mole ratio of pentane $(A)$ and hexane $(B)$ is $1:4$. At $20^o C$,the vapor pressures of these pure hydrocarbons are $440 \ mm$ and $120 \ mm$ respectively. The mole fraction of pentane in the vapor phase is.....

The vapour pressure of water at $20\, ^oC$ is $17.5\, mm\, Hg$. If $18\, g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2\, g$ of water at $20\, ^oC$,the vapour pressure of the resulting solution will be .......... $mm\, Hg$.