If two substances $A$ and $B$ have $P_{A}^{\circ} : P_{B}^{\circ} = 1 : 2$ and have mole fraction in solution in the ratio $1 : 2$,then mole fraction of $A$ in vapour phase is:-

What is the vapour pressure of a solution when $2 \ mol$ of a non-volatile solute are dissolved in $20 \ mol$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

Which of the following changes will cause an increase in the vapour pressure of a $1 \ m$ aqueous $KI$ solution at the same temperature?

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 $20^\circ C$,what will be the vapour pressure of the resulting solution (in $mm \ Hg$)?

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:

The vapour pressure of a pure solvent is $0.70 \ atm.$ When a non-volatile solute $A$ is dissolved in it,the vapour pressure of the solution becomes $0.35 \ atm.$ The mole fraction of solute $A$ will be$:-$