Calculate the vapour pressure of a solution containing a mixture of $2 \ mol$ of volatile liquid $A$ and $3 \ mol$ of volatile liquid $B$ at room temperature. $(P_{A}^{\circ} = 420 \ mm \ Hg, P_{B}^{\circ} = 610 \ mm \ Hg)$ (in $mm \ Hg$)

If two substances $A$ and $B$ have $P_A^0 : P_B^0 = 1 : 2$ and have mole fraction in solution $1 : 2$,then what is the mole fraction of $A$ in the vapour phase?

Relative lowering of vapour pressure of a dilute solution is $0.5$. What is the mole fraction of the non-volatile solute?

Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^o : P_B^o = 1 : 3$ at a certain temperature. If the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4 : 3$,then the mole fraction of $B$ in the solution at the same temperature is

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 120 - 80 \, X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and $B$ at the same temperature are respectively:

The vapour pressure of a given liquid decreases if?