Lowering of vapour pressure due to a solute in $1 \, molal$ aqueous solution at $100 \, ^\circ C$ is ........ $torr$.

The relative lowering in vapour pressure of an aqueous solution of urea which is $6\%$ by weight,is

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 $180 \ g \ mol^{-1}$

If vapour pressure of pure solvent and solution are $240 \ mm Hg$ and $216 \ mm Hg$ respectively,then the mole fraction of solvent in the solution is:

$A$ 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 $\text{ mm Hg}$)

Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene having mole fraction of toluene $0.50$. If vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature,mole fraction of toluene in vapour phase will be: