Calculate the relative lowering of vapour pressure if the vapour pressure of benzene and vapour pressure of solution of non-volatile solute in benzene are $640 \ mmHg$ and $590 \ mmHg$ respectively at the same temperature.

Vapour pressure of pure water at $298 \, K$ is $23.8 \, mm \, Hg$. $50 \, g$ of urea $(NH_2CONH_2)$ is dissolved in $850 \, g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

What is the vapour pressure of a solution containing $0.1 \text{ mol}$ of non-volatile solute dissolved in $16.2 \text{ g}$ water (in $\text{ mmHg}$)? ($P_1^0 = 24 \text{ mmHg}$, molar mass of water $18 \text{ g mol}^{-1}$)

Vapour pressure in $mm \ Hg$ of $0.1 \ mole$ of urea in $180 \ g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 \ mm \ Hg$ )

In an experiment,$1 \ g$ of a non-volatile solute was dissolved in $100 \ g$ of acetone (molar mass $= 58 \ g/mol$) at $298 \ K$. The vapour pressure of the solution was found to be $192.5 \ mm \ Hg$. The molecular weight of the solute is (vapour pressure of pure acetone $= 195 \ mm \ Hg$). (in $.24$)

$A$ 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.