The vapor pressures of two pure liquids $(A)$ and $(B)$ are $100 \ Torr$ and $80 \ Torr$ respectively. What is the total vapor pressure of the solution formed by mixing $2 \ moles$ of $(A)$ and $3 \ moles$ of $(B)$ (in $Torr$)?

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 relative lowering of vapour pressure produced by dissolving $71.5 \ g$ of a substance in $1000 \ g$ of water is $0.00713$. The molecular weight of the substance will be:

Calculate the amount of solute dissolved in $612 \ g$ of water at $30^{\circ} C$ if molar mass of solute is $342 \ g \ mol^{-1}$ (Relative vapour pressure lowering is $0.025$ and molar mass of water $18 \ g \ mol^{-1}$ ). (in $g$)

The boiling points of $C_6H_6, CH_3OH, C_6H_5NH_2$ and $C_6H_5NO_2$ are $80 \ ^oC, 65 \ ^oC, 184 \ ^oC$ and $212 \ ^oC$ respectively. Which will show the highest vapour pressure at room temperature?

$A$ solution is made by mixing $1 \ mole$ of volatile liquid $A$ with $3 \ moles$ of volatile liquid $B$. The vapour pressure of pure $A$ is $200 \ mm \ Hg$ and that of the solution is $500 \ mm \ Hg$. The vapour pressure of pure $B$ and the least volatile component of the solution,respectively,are $:$