$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

At $30^\circ C$,liquid $A$ and $B$ form an ideal solution. The total vapor pressure of a solution containing $1 \ mol$ of $A$ and $2 \ mol$ of $B$ is $250 \ mm \ Hg$. The total vapor pressure becomes $300 \ mm \ Hg$ when $1 \ mol$ more of $A$ is added to the first solution. What are the vapor pressures of pure $A$ and $B$ at the same temperature?

Vapour pressure of pure water at $298 \, K$ is $23.8 \, mm \, Hg$. $50 \, g$ of urea $(NH_2CONH_2)$ is dissolved in $850 \, g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

At $80\,^oC,$ the vapour pressure of pure liquid $'A'$ is $520\, mm\,Hg$ and that of pure liquid $'B'$ is $1000\, mm\,Hg.$ If a mixture solution of $'A'$ and $'B'$ boils at $80\,^oC$ and $1\, atm$ pressure,the amount of $'A'$ in the mixture is ........... $mol$ percent $(1\, atm = 760\, mm\,Hg).$

If $p^{\circ}$ and $p$ are the vapour pressure of the pure solvent and solution, and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution, then the correct relation between $p$ and $p^{\circ}$ is:

An aqueous solution is $1.00 \ molal$ in $KI$. Which change will cause the vapour pressure of the solution to increase?