What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water $\left(P_1^0 = 32 \ mm \ Hg\right) (in $mm \ Hg$)?$

$A$ solution of liquids $A$ and $B$ is given. The mole fraction of $A$ in the vapor phase is $x_1$ and in the liquid phase is $x_2$. If $P_A^o$ and $P_B^o$ are the vapor pressures of pure $A$ and $B$ respectively,then the total vapor pressure $P_{total}$ will be:

$A$ liquid is kept in a closed vessel. If a glass plate (negligible mass) with a small hole is kept on top of the liquid surface,then the vapour pressure of the liquid in the vessel is

Calculate the vapour pressure of pure volatile liquid $B$ $(P_B^o)$ at a given temperature if the mole fraction of liquid $B$ $(x_B)$ is $0.4$ and the vapour pressure of pure volatile liquid $A$ $(P_A^o)$ is $400 \ mm \ Hg$. The total vapour pressure of the solution is $P_{total} = 600 \ mm \ Hg$. (in $mm \ Hg$)

Using Raoult's law,explain how the total vapour pressure over the solution is related to the mole fraction of components in the following solutions:
$(a)$ $CHCl_{3(l)}$ and $CH_{2}Cl_{2(l)}$
$(b)$ $NaCl_{(s)}$ and $H_{2}O_{(l)}$

$6 \ g$ of a non-volatile solute $(x)$ is dissolved in $100 \ g$ of water. The relative lowering of vapour pressure of the resultant solution is $0.006$. What is the molar mass (in $g \ mol^{-1}$) of $x$?