At a certain temperature,the mole fraction of a solute is $0.15$ and the vapor pressure of the pure solvent is $120 \, torr$. If the solute is non-volatile,what will be the relative lowering in vapor pressure?

$A$ solution is prepared by mixing $8.5 \ g$ of $CH_2Cl_2$ and $11.95 \ g$ of $CHCl_3$. If the vapour pressures of pure $CH_2Cl_2$ and $CHCl_3$ at $298 \ K$ are $415 \ mm \ Hg$ and $200 \ mm \ Hg$ respectively,the mole fraction of $CHCl_3$ in the vapour phase is: (Molar mass of $Cl = 35.5 \ g \ mol^{-1}$)

At $T \ K$, $0.1 \ mol$ of a non-volatile solute was dissolved in $0.9 \ mol$ of a volatile solvent. The vapour pressure of pure solvent is $0.9 \ bar$. What is the vapour pressure (in $bar$) of the solution?

The vapour pressure of pure liquid solvent is $0.50 \ atm$. When a non-volatile solute $B$ is added to the solvent,its vapour pressure drops to $0.30 \ atm$. The mole fraction of the solute $B$ is $......$

Vapour pressure of solvent $A$ is $0.90 \text{ atm}$. When a non-volatile solute is added,the vapour pressure drops to $0.60 \text{ atm}$. What is the mole fraction of $A$ in the solution?

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 $CH_2Cl_2 = 85 \ u$)