When a non-volatile solute is added to the solvent,the vapour pressure of the solvent decreases by $10 \ mm \ Hg$. The mole fraction of the solute in the solution is $0.2$. What would be the mole fraction of the solvent if the decrease in vapour pressure is $20 \ mm \ Hg$?

$12 \ g$ of a nonvolatile solute dissolved in $108 \ g$ of water produces the relative lowering of vapour pressure of $0.1$. The molecular mass of the solute is

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).

Assertion : If a liquid solute more volatile than the solvent is added to the solvent,the vapour pressure of the solution may increase i.e.,$p_s > p^o$.
Reason : In the presence of a more volatile liquid solute,only the solute will form the vapours and solvent will not.

Dry air is passed through a solution containing $10 \ g$ of solute in $90 \ g$ of water and then through pure water. The loss in weight of the solution is $2.5 \ g$ and the loss in weight of the solvent is $0.05 \ g$. The molecular weight of the solute is:

At $350 \ K$,the vapour pressure of pure liquids $A$ and $B$ are $450 \ mm \ Hg$ and $700 \ mm \ Hg$ respectively. If the total vapour pressure of the liquid mixture is $600 \ mm \ Hg$,the mole fractions of $A$ and $B$ in the vapour phase respectively are: