Find the molar mass of a nonvolatile solute when $20 \ g$ of it is dissolved in $200 \ g$ of water at $300 \ K$. [Relative lowering of vapour pressure $= 0.02$]

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:

$651 \ g$ of ethylene glycol $(HOCH_2CH_2OH)$ is dissolved in $1.5 \ kg$ of water at $363 \ K$. The vapour pressure of pure water at $363 \ K$ is $0.7 \ atm$. Assuming ideal solution behaviour,the vapour pressure of water over the solution (in $atm$) is closest to $...$

$6 \ g$ of a non-volatile solute $(x)$ is dissolved in $100 \ g$ of water. The relative lowering of vapour pressure of the resultant solution is $0.006$. What is the molar mass (in $g \ mol^{-1}$) of $x$?

Two liquids $A$ and $B$ form an ideal solution. At $320 \ K$, the vapour pressure of the solution, containing $3 \ mol$ of $A$ and $1 \ mol$ of $B$ is $500 \ mm \ Hg$. At the same temperature, if $1 \ mol$ of $A$ is further added to this solution, the vapour pressure of the solution increases by $20 \ mm \ Hg$. The vapour pressure (in $mm \ Hg$) of $B$ in the pure state is . . . . . . (Nearest integer).

Lowering of vapour pressure is highest for