What is the vapour pressure of a solution when $2 \ mol$ of a non-volatile solute are dissolved in $20 \ mol$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

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:

Calculate the vapour pressure of the solution if the relative lowering of vapour pressure and the vapour pressure of the pure solvent are $0.018$ and $18 \ mm \ Hg$ respectively at $300 \ K$. (in $mm \ Hg$)

Among acetone,ether,and ethanol,which one will have the maximum and which one will have the minimum vapor pressure?

$A$ solution was prepared by dissolving $0.1 \ mol$ of a nonvolatile solute in $0.9 \ mol$ of water. What is the relative lowering of vapour pressure of the solution?

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 $Hg$?