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:

The vapour pressure of a solvent decreases by $20 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.5$. What should be the mole fraction of the solvent for the decrease in the vapour pressure to be $10 \ mm$ of $Hg$?

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 $20 \ mm \ Hg$?

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.

At $80\,^{\circ}C$,the vapour pressure of pure liquid '$A$' is $520\, mm\, Hg$ and that of pure liquid '$B$' is $1000\, mm\, Hg$. If a mixture solution of '$A$' and '$B$' boils at $80\,^{\circ}C$ and $1\, atm$ pressure,then the amount of '$A$' in the mixture is ......... $mol\, \%$. $(1\, atm = 760\, mm\, Hg)$

The variation of vapour pressure $(b)$ as a function of temperature $(a)$ is studied for $C_2H_5OC_2H_5$,$CCl_4$,and $H_2O$ at $760 \ mm \ Hg$ and is shown in the figure below. The boiling temperatures of $C_2H_5OC_2H_5$,$CCl_4$,and $H_2O$ are $308 \ K$,$350 \ K$,and $373 \ K$ respectively. Curves $A$,$B$,and $C$ respectively correspond to: