The vapor pressure of a pure liquid solvent $(X)$ decreases from $0.80 \ atm$ to $0.60 \ atm$ upon the addition of a non-volatile solute $(Y)$. What is the mole fraction of $(Y)$ in the solution?

What is the mass in $g$ of a non-volatile solute with a molecular weight of $40$ that should be dissolved in $57 \ g$ of octane to reduce its vapour pressure to $80 \%$ of its original value?

What weight of glucose must be dissolved in $100 \ g$ of water to lower the vapour pressure by $0.20 \ mm \ Hg$ (in $g$)? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is $54.2 \ mm \ Hg$ at room temperature. Molar mass of glucose is $180 \ g \ mol^{-1}$

At $20^{\circ} C$,the vapour pressure of benzene is $70 \, torr$ and that of methyl benzene is $20 \, torr$. The mole fraction of benzene in the vapor phase at $20^{\circ} C$ above an equimolar mixture of benzene and methyl benzene is $..... \times 10^{-2}$ (Nearest integer).

If the vapor pressure of a solution formed by dissolving $1 \ mol$ of a non-volatile solute in $4 \ mol$ of a solvent is $24.0 \ kPa$,what will be the vapor pressure of the pure solvent in $kPa$?

$34.2 \ g$ of cane sugar is dissolved in $180 \ g$ of water. The relative lowering of vapour pressure will be