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 $Torr$)?

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:

The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600\, mm\, Hg$,respectively at $298\, K$. On mixing the two liquids,the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid $B$ is $0.5$ in the mixture. The vapour pressure of the final solution,the mole fractions of components $A$ and $B$ in vapour phase,respectively are

The vapour pressure of pure water is $92.5 \ mm$ at $300 \ K$. The vapour pressure of $1 \ molal$ solution of $X$ in water is :-

Two liquids $A$ and $B$ form an ideal solution. At $320 \ K$, the vapour pressure of the solution, containing $3 \ mol$ of $A$ and $1 \ mol$ of $B$ is $500 \ mm \ Hg$. At the same temperature, if $1 \ mol$ of $A$ is further added to this solution, the vapour pressure of the solution increases by $20 \ mm \ Hg$. The vapour pressure (in $mm \ Hg$) of $B$ in the pure state is . . . . . . (Nearest integer).

$3 \ g$ of urea is dissolved in $45 \ g$ of $H_2O$. The relative lowering in vapour pressure is