$1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ is mixed with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$. The vapor pressure of the resulting ideal solution is .......... $mm \ Hg$.

The vapour pressure of pure benzene and toluene are $160 \ torr$ and $60 \ torr$ respectively. The mole fraction of toluene in the vapour phase in contact with an equimolar solution of benzene and toluene is:

Calculate the molar mass of a nonvolatile solute if $6.4 \text{ g}$ of it dissolved in $100 \text{ g}$ water produces a relative lowering in vapour pressure of $0.016$ at $300 \text{ K}$.

The relative lowering of vapor pressure is equal to the mole fraction of the non-volatile solute. This statement was given by whom?

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?

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 .......... $torr$.