Two liquids $A$ and $B$ have $P_A^o$ and $P_B^o$ in the ratio of $1 : 3$ and the ratio of number of moles of $A$ and $B$ in the liquid phase is $1 : 3$. The mole fraction of $A$ in the vapour phase in equilibrium with the solution is equal to:

The vapour pressure of pure $CHCl_3$ and $CH_2Cl_2$ are $200 \,atm$ and $41.5 \,atm$ respectively. The weights of $CHCl_3$ and $CH_2Cl_2$ are respectively $11.9 \,g$ and $17 \,g$. The vapour pressure of the solution will be (in $,atm$)

Assertion : If a liquid solute more volatile than the solvent is added to the solvent,the vapour pressure of the solution may increase i.e.,$p_s > p^o$.
Reason : In the presence of a more volatile liquid solute,only the solute will form the vapours and solvent will not.

The relative lowering of vapour pressure produced by dissolving $71.5 \ g$ of a substance in $1000 \ g$ of water is $0.00713$. The molecular weight of the substance will be:

$A$ 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:

The plot of total vapour pressure as a function of mole fraction of the components of an ideal solution formed by mixing liquids $X$ and $Y$ is