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:

Two liquids $X$ and $Y$ form an ideal solution. At a constant temperature of $300 \ K$,the vapor pressure of a solution containing $1 \ mol$ of $X$ and $3 \ mol$ of $Y$ is $550 \ mm \ Hg$. If $1 \ mol$ of $Y$ is added to this solution,the vapor pressure of the solution increases by $10 \ mm \ Hg$. What are the vapor pressures of $X$ and $Y$ in their pure states?

$A$ mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290 \, mm$ at $300 \, K$. The vapour pressure of propyl alcohol is $200 \, mm$. If the mole fraction of ethyl alcohol is $0.6$,its vapour pressure (in $mm$) at the same temperature will be:

The vapour pressure of two pure liquids $(A)$ and $(B)$ are $100 \ torr$ and $80 \ torr$ respectively. The total pressure of the solution obtained by mixing $2 \ mole$ of $(A)$ and $3 \ mole$ of $(B)$ would be ........ $torr$.

An aqueous solution of $2 \%$ non-volatile solute exerts a pressure of $1.004 \ bar$ at the normal boiling point of the solvent. What is the molar mass of the solute?

At $T \ K$,$0.1 \ mol$ of a non-volatile solute was dissolved in $0.9 \ mol$ of a volatile solvent. The vapour pressure of pure solvent is $0.9 \ bar$. What is the vapour pressure (in $bar$) of the solution?