$12 \ g$ 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

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 number of moles of chlorobenzene and bromobenzene are $0.1$ and $0.2$ respectively. The vapour pressure of pure chlorobenzene and pure bromobenzene are $0.350 \ bar$ and $0.500 \ bar$ respectively. Calculate the total vapour pressure of the solution prepared by mixing these two components.

$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.

The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

Lowering of vapour pressure is highest for