In a solution,the mole fraction of solute is $0.2$ when the lowering in vapour pressure is $10 \text{ mm Hg}$. To achieve a lowering of vapour pressure of $20 \text{ mm Hg}$,the mole fraction of the solute in the solution should be:

At room temperature,a dilute solution of urea is prepared by dissolving $0.60 \ g$ of urea in $360 \ g$ of water. If the vapour pressure of pure water at this temperature is $35 \ mm \ Hg$,the lowering of vapour pressure will be: .............. $mm \ Hg$ (molar mass of urea $= 60 \ g \ mol^{-1}$)

The vapour pressures of $A$ and $B$ at $25^{\circ} C$ are $90 \ mm \ Hg$ and $15 \ mm \ Hg$ respectively. If $A$ and $B$ are mixed such that the mole fraction of $A$ in the mixture is $0.6$,then the mole fraction of $B$ in the vapour phase is $x \times 10^{-1}$. The value of $x$ is $.....$ (Nearest integer)

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

$A$ gaseous mixture of two substances $A$ and $B$,under a total pressure of $0.8 \ atm$ is in equilibrium with an ideal liquid solution. The mole fraction of substance $A$ is $0.5$ in the vapour phase and $0.2$ in the liquid phase. The vapour pressure of pure liquid $A$ is $...... \ atm$. (Nearest integer)

The vapour pressure of a dilute aqueous solution of glucose is $750 \ mm$ of mercury at $373 \ K$. The mole fraction of solvent is