At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be:

$3 \ g$ of urea is dissolved in $45 \ g$ of $H_2O$. The relative lowering in vapour pressure is

Vapour pressure of $CCl_4$ at $25 \text{ }^\circ\text{C}$ is $143 \text{ mm Hg}$. If $0.5 \text{ g}$ of a non-volatile solute is dissolved in $100 \text{ cm}^3$ of $CCl_4$, find the vapour pressure of the solution. (Density of $CCl_4 = 1.58 \text{ g/cm}^3$ and molar mass of solute = $65 \text{ g/mol}$) (in $\text{ mm Hg}$)

$3 \ g$ of urea is dissolved in $45 \ g$ of water. The relative lowering of vapour pressure is

At a given temperature,the vapour pressure of pure water is $25 \ mm$ and the vapour pressure of a dilute solution of urea is $24.5 \ mm$. The molality of the solution is:

The vapor pressures of two pure liquids $(A)$ and $(B)$ are $100 \ Torr$ and $80 \ Torr$ respectively. What is the total vapor pressure of the solution formed by mixing $2 \ moles$ of $(A)$ and $3 \ moles$ of $(B)$ (in $Torr$)?