The vapor pressure of a liquid mixture of $X$ and $Y$ is given by the equation $P = 160X_x + 50$,where $X_x$ is the mole fraction of $X$. The ratio of the vapor pressures of pure liquids $X$ and $Y$ is:

The graph between vapour pressure and mole fraction of the solvent for an ideal solution is:

Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene having mole fraction of toluene $0.50$. If vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature,mole fraction of toluene in vapour phase will be:

At $300 \ K$,the vapour pressures of two pure liquids,$A$ and $B$ are $100 \ mm \ Hg$ and $500 \ mm \ Hg$,respectively. If in a mixture of $A$ and $B$,the total vapour pressure is $300 \ mm \ Hg$,the mole fractions of $A$ in the liquid and in the vapour phase,respectively,are:

At $25\,^oC$,the vapour pressure of pure liquid $A$ (mol. wt. $= 40$) is $100\,torr$,while that of pure liquid $B$ is $40\,torr$ (mol. wt. $= 80$). The vapour pressure at $25\,^oC$ of a solution containing $20\,g$ of each $A$ and $B$ is ........... $torr$.

Which of the following changes will cause an increase in the vapour pressure of a $1 \ m$ aqueous $KI$ solution at the same temperature?