At $298 \ K$,$0.714$ moles of liquid $A$ is dissolved in $5.555$ moles of liquid $B$. The vapour pressure of the resultant solution is $475 \ torr$. The vapour pressure of pure liquid $A$ at the same temperature is $280.7 \ torr$. What is the vapour pressure of pure liquid $B$ in $torr$?

At $T$ $(K)$,the vapour pressures of pure liquids $A$ and $B$ are $100 \ mm$ and $160 \ mm$ respectively. An ideal solution is formed by mixing $2 \ moles$ of $A$ and $3 \ moles$ of $B$ at the same temperature. The mole fractions of $A$ and $B$ in the vapour state respectively are

$X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ at different concentrations:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is:

$2\, \text{moles}$ of liquid $A$ is mixed with $3\, \text{moles}$ of liquid $B$. Calculate the mole fraction of $A$ in the vapour phase if $P_A^o = 100\, \text{mm}$ and $P_B^o = 200\, \text{mm}$.

The vapour pressure of pure liquids $A$ and $B$ are $450 \ mm$ and $700 \ mm$ of $Hg$ at $350 \ K$ respectively. If the total vapour pressure of the mixture is $600 \ mm$ of $Hg$,the composition of the mixture in the solution is

Find the molar mass of solute when $2 \ g$ is dissolved in $60 \ g$ of benzene,and the relative lowering of vapour pressure is $0.06$. (Molar mass of benzene is $78 \ g \ mol^{-1}$)