At $88\,^oC$,the vapor pressure of benzene is $900 \, \text{torr}$ and that of toluene is $360 \, \text{torr}$. What is the mole fraction of benzene in the mixture with toluene that will boil at $88\,^oC$ and $1 \, \text{atm}$ pressure,assuming it forms an ideal solution?

$A$ solution of two miscible liquids showing negative deviation from Raoult's law will have :

Which of the following liquid mixtures shows a positive deviation from Raoult's law?

Two liquids $A$ and $B$ form an ideal solution. What will be the vapor pressure of a solution containing a mole ratio of $A$ and $B$ as $3:1$ in $torr$? $(Given: P_A^o = 24 \ torr, P_B^o = 40 \ torr)$

Two liquids $X$ and $Y$ form an ideal solution. At $300 \ K,$ vapour pressure of the solution containing $1 \ mol$ of $X$ and $3 \ mol$ of $Y$ is $550 \ mm \ Hg.$ At the same temperature,if $1 \ mol$ of $Y$ is further added to this solution,vapour pressure of the solution increases by $10 \ mm \ Hg.$ Vapour pressure (in $mm \ Hg$) of $X$ and $Y$ in their pure states will be,respectively:

Which property is shown by an ideal solution?