Calculate the relative lowering of vapour pressure if the vapour pressure of benzene and vapour pressure of solution of non-volatile solute in benzene are $640 \ mmHg$ and $590 \ mmHg$ respectively at the same temperature.

"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

At $300 \ K$,the vapor pressures of two pure liquids $A$ and $B$ are $150 \ mm \ Hg$ and $100 \ mm \ Hg$,respectively. If the mole fractions of $A$ and $B$ in the solution are equal,then the mole fraction of $B$ in the vapor phase at the same temperature is:

When a non-volatile solute is added to a pure solvent,the vapor pressure decreases by $11.5 \ torr$. If the mole fraction of the solute is $0.2$,what will be the vapor pressure of the pure solvent in $torr$?

$A$ solution of a non-volatile solute is obtained by dissolving $2 \ g$ of solute in $50 \ g$ of benzene. Calculate the vapour pressure of the solution if the vapour pressure of pure benzene is $640 \ mmHg$ at $25^{\circ} C$. [Molar mass of benzene $= 78 \ g \ mol^{-1}$,Molar mass of solute $= 64 \ g \ mol^{-1}$] (in $mm \ Hg$)

Vapour pressure of chloroform $(CHCl_3)$ and dichloromethane $(CH_2Cl_2)$ at $25^\circ C$ are $200 \ mm \ Hg$ and $41.5 \ mm \ Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \ g$ of $CHCl_3$ and $40 \ g$ of $CH_2Cl_2$ at the same temperature will be (Molecular mass of $CHCl_3 = 119.5 \ u$ and molecular mass of $CH_2Cl_2 = 85 \ u$)