The vapour pressure of two liquids $P$ and $Q$ are $80$ and $60 \, torr$,respectively. The total vapour pressure of the solution obtained by mixing $3 \, mole$ of $P$ and $2 \, mole$ of $Q$ would be .......... $torr$.

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 $mm \ Hg$)

The vapour pressure of benzene at a certain temperature is $640 \, mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \, g$ is added to $39.08 \, g$ of benzene. The vapour pressure of the solution is $600 \, mm$ of $Hg$. What is the molecular weight of the solid substance?

For a dilute solution,Raoult's law states that

Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

What weight of glucose must be dissolved in $100 \ g$ of water to lower the vapour pressure by $0.20 \ mm \ Hg$ (in $g$)? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is $54.2 \ mm \ Hg$ at room temperature. Molar mass of glucose is $180 \ g \ mol^{-1}$