The mass of non-volatile,non-electrolyte solute (molar mass $= 50 \ g \ mol^{-1}$) needed to be dissolved in $114 \ g$ octane to reduce its vapour pressure to $75\%$ is .............. $g$.

At $25\,^{\circ}C$,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 $25\,^{\circ}C$ of a solution containing $20\, g$ of each $A$ and $B$ is .......... $torr$.

$A$ solution of a non-volatile solute is obtained by dissolving $2 \ g$ of solute in $50 \ g$ of benzene. Calculate the vapour pressure of the solution if the vapour pressure of pure benzene is $640 \ mmHg$ at $25^{\circ} C$. [Molar mass of benzene $= 78 \ g \ mol^{-1}$,Molar mass of solute $= 64 \ g \ mol^{-1}$] (in $mm \ Hg$)

The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be

The vapour pressure of water at room temperature is lowered by $5\%$ by dissolving a solute in it. Then the approximate molality of the solution is

Lowering of vapour pressure due to a solute in $1 \, molal$ aqueous solution at $100 \, ^\circ C$ is ........ $torr$.