What is the vapour pressure of a solution containing $0.1 \text{ mol}$ of non-volatile solute dissolved in $16.2 \text{ g}$ water (in $\text{ mmHg}$)? ($P_1^0 = 24 \text{ mmHg}$, molar mass of water $18 \text{ g mol}^{-1}$)

$A$ solution is prepared by dissolving $68 \text{ g}$ of sucrose $(C_{12}H_{22}O_{11})$ in $1 \text{ kg}$ of water. Calculate the vapour pressure of the solution at $298 \text{ K}$. Given: Vapour pressure of pure water at $298 \text{ K} = 18 \text{ mm Hg}$ and mole fraction of solvent $(x_1)$ $= 0.9964$. (in $\text{ mm Hg}$)

For two components $A$ and $B$,$P_A^0 : P_B^0 = 1:2$ and $X_A : X_B = 1:2$. The mole fraction of $A$ in the vapor phase is:

Which of the following solutions in water will have the lowest vapour pressure?

The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600\, mm\, Hg$,respectively at $298\, K$. On mixing the two liquids,the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid $B$ is $0.5$ in the mixture. The vapour pressure of the final solution,the mole fractions of components $A$ and $B$ in vapour phase,respectively are

What is the molar mass of a solute when $2.3 \ g$ of a non-volatile solute is dissolved in $46 \ g$ of benzene at $30^{\circ} C$? (Relative lowering of vapour pressure is $0.06$ and the molar mass of benzene is $78 \ g \ mol^{-1}$)