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:

What will be the molar mass of a non-volatile solute if the vapour pressure of pure benzene is $450 \ mm \ Hg$ and it decreases to $400 \ mm \ Hg$ when $1.5 \ g$ of the solute is added to $30 \ g$ of benzene? (Atomic mass: $C=12, H=1$)

The vapour pressure of water is $12.3 \, kPa$ at $300 \, K$. Calculate the vapour pressure of a $1 \, molal$ solution of a non-volatile solute in it.

The vapor pressure of a liquid mixture of $X$ and $Y$ is given by the equation $P = 160X_x + 50$,where $X_x$ is the mole fraction of $X$. The ratio of the vapor pressures of pure liquids $X$ and $Y$ is:

An aqueous solution containing $28\%$ by mass of a liquid $A$ (molar mass $= 140 \ g/mol$) has a vapour pressure of $160 \ mm$ at $37 \ ^\circ C$. Find the vapour pressure of the pure liquid $A$ approximately .......... $mm$. (The vapour pressure of pure water at $37 \ ^\circ C$ is $150 \ mm$)

When a certain amount of solid $A$ is dissolved in $100 \ g$ of water at $25^{\circ} \ C$ to make a dilute solution,the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $23.76 \ mm \ Hg$. The number of moles of solute $A$ added is $...........$ (Nearest Integer).