The vapour pressure will be lowest for

Calculate the vapour pressure of a solution containing a mixture of $2 \ moles$ of volatile liquid $A$ and $3 \ moles$ of volatile liquid $B$ at room temperature. $(P_{A}^{\circ} = 420 \ mm \ Hg, P_{B}^{\circ} = 610 \ mm \ Hg)$ (in $mm \ Hg$)

At $T \ K$,the vapour pressure of an aqueous solution of a non-volatile solute,whose mole fraction is $0.02$,is found to be $34.65 \ mm \ Hg$. What is the vapour pressure (in $mm \ Hg$) of pure water at the same temperature?

Calculate the relative lowering of vapour pressure if the vapour pressure of benzene and vapour pressure of solution of non-volatile solute in benzene are $640 \ mmHg$ and $590 \ mmHg$ respectively at the same temperature.

Vapour pressure of solvent $A$ is $0.90 \text{ atm}$. When a non-volatile solute is added,the vapour pressure drops to $0.60 \text{ atm}$. What is the mole fraction of $A$ in the solution?

The vapour pressure of pure liquid solvent is $0.50 \ atm$. When a non-volatile solute $B$ is added to the solvent,its vapour pressure drops to $0.30 \ atm$. The mole fraction of the solute $B$ is $......$