The vapour pressure of a pure liquid $A$ is $70 \ torr$ at $300 \ K$. It forms an ideal solution with another liquid $B$. The mole fraction of $B$ is $0.2$ and the total vapour pressure of the solution is $84 \ torr$ at the same temperature. The vapour pressure of pure liquid $B$ (in $torr$) is:

$A$ solution is prepared by dissolving $10 \ g$ of a non-volatile solute (molar mass,'$M$ $g \ mol^{-1}$') in $360 \ g$ of water. What is the molar mass in $g \ mol^{-1}$ of solute if the relative lowering of vapour pressure of solution is $5 \times 10^{-3}$?

At $300 \ K$,the vapour pressure of a solution containing $1 \ mole$ of $n-hexane$ and $3 \ moles$ of $n-heptane$ is $550 \ mm \ Hg$. At the same temperature,if one more mole of $n-heptane$ is added to this solution,the vapour pressure of the solution increases by $10 \ mm \ Hg$. What is the vapour pressure in $mm \ Hg$ of $n-heptane$ in its pure state?

$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

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

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 $Torr$)?