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 $mm \ Hg$)

$x \ g$ of a solute is dissolved in $y \ g$ of two different liquids $A$ and $B$. The relative lowering of vapor pressure of the solution in $A$ is twice that of the solution in $B$. If the moles of solute are negligible compared to the moles of solvent,which of the following statements regarding the molar masses $M_A$ and $M_B$ of the solvents is correct?

$A$ solution contains $KCl$ in water. It is heated continuously and the vapour formed is removed continuously. The vapour pressure of the solution will be:

What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water $\left(P_1^0 = 32 \ mm \ Hg\right) (in $mm \ Hg$)?$

When a non-volatile solute is added to a pure solvent,the vapor pressure decreases by $11.5 \ torr$. If the mole fraction of the solute is $0.2$,what will be the vapor pressure of the pure solvent in $torr$?

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