At $T$ $(K)$,the vapour pressures of pure liquids $A$ and $B$ are $100 \ mm$ and $160 \ mm$ respectively. An ideal solution is formed by mixing $2 \ moles$ of $A$ and $3 \ moles$ of $B$ at the same temperature. The mole fractions of $A$ and $B$ in the vapour state respectively are

Calculate the relative lowering of vapour pressure if the vapour pressure of the pure solvent and the vapour pressure of the solution at $25^{\circ} C$ are $32 \ mm \ Hg$ and $30 \ mm \ Hg$ respectively.

Vapour pressure of chloroform $(CHCl_{3})$ and dichloromethane $(CH_{2}Cl_{2})$ at $298\,K$ are $200\,mm\,Hg$ and $415\,mm\,Hg$ respectively. $(i)$ Calculate the vapour pressure of the solution prepared by mixing $25.5\,g$ of $CHCl_{3}$ and $40\,g$ of $CH_{2}Cl_{2}$ at $298\,K$ and,$(ii)$ mole fractions of each component in vapour phase.

The vapour pressure of water at $20\,^{\circ}C$ is $17.5\, mm\, Hg.$ If $18\, g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2\, g$ of water at $20\,^{\circ}C,$ the vapour pressure of the resulting solution will be $.........\, mm$ of $Hg$.

Two $5 \ molal$ solutions are prepared by dissolving a non-electrolyte,non-volatile solute separately in the solvents $X$ and $Y$. The molecular weights of the solvents are $M_X$ and $M_Y$,respectively,where $M_X = \frac{3}{4} M_Y$. The relative lowering of vapour pressure of the solution in $X$ is $m$ times that of the solution in $Y$. Given that the number of moles of solute is very small in comparison to that of solvent,the value of $m$ is

Vapour pressure of a pure solvent is $550 \ mm$ of $Hg$. By addition of a non-volatile solute,it decreases to $510 \ mm$ of $Hg$. Calculate the mole fraction of the solute in the solution.