Calculate the relative lowering of vapour pressure of a solution containing $0.56 \ g$ of a non-volatile solute in $100 \ g$ of water [molar mass of solute $= 60 \ g \ mol^{-1}$].

For a solution of volatile liquids,the partial vapour pressure of each component in the solution is directly proportional to:

$A$ 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 $Cl = 35.5 \ g \ mol^{-1}$)

Vapour pressure of $CCl_4$ at $25 \text{ }^\circ\text{C}$ is $143 \text{ mm Hg}$. If $0.5 \text{ g}$ of a non-volatile solute is dissolved in $100 \text{ cm}^3$ of $CCl_4$, find the vapour pressure of the solution. (Density of $CCl_4 = 1.58 \text{ g/cm}^3$ and molar mass of solute = $65 \text{ g/mol}$) (in $\text{ mm Hg}$)

At $T \ K$,the vapour pressure of an $x$ molal aqueous solution containing a non-volatile solute is $12.078 \ kPa$. The vapour pressure of pure water at $T \ K$ is $12.3 \ kPa$. What is the value of $x$?

The vapour pressure lowering caused by the addition of $100 \, g$ of sucrose (molecular mass $= 342$) to $1000 \, g$ of water if the vapour pressure of pure water at $25 \, ^oC$ is $23.8 \, mm \, Hg$.