$0.5 \mathrm{g}$ of a non-volatile organic compound (Molar weight=$65$) is dissolved in $100 \mathrm{mL}$ of ${\mathrm{CCl}}_4.$ If the vapour pressure of pure ${\mathrm{CCl}}_4$ is $143 \mathrm{mm} \mathrm{Hg}$, what would be the vapour pressure of the solution? (Density of ${\mathrm{CCl}}_4$ solution is $1.58 \mathrm{g}/\mathrm{cc}$.)

Which of the following aqueous solutions has a higher vapour pressure if the density of water is $1 \mathrm{g}/\mathrm{cc}?$

(a) Solution having mole fraction of cane sugar=$0.1$

(b) Solution having molal concentration=$1 \mathrm{m}$

At $25° \mathrm{C}$, benzene and toluene have densities $0.879$ and $0.867 \mathrm{g}/\mathrm{cc}$, respectively. Assuming that benzene-toluene solutions are ideal, establish the equation for the density of the solution:

$\mathrm{d}=\frac{1}{100}[0·879\mathrm{V}+0.867(100-\mathrm{V}\left)\right]$

where $\mathrm{V}$ is the volume per cent of benzene.

Ethanol and methanol form a solution that is very nearly ideal. The vapour pressure of ethanol is $44.5 \mathrm{mm}$ and that of methanol is $88.7 \mathrm{mm}$ at $20° \mathrm{C}$.

Calculate the mole fraction of methanol and ethanol in a solution obtained by mixing $60 \mathrm{g}$of ethanol with $40 \mathrm{g}$ of methanol.

Ethanol and methanol form a solution that is very nearly ideal. The vapour pressure of ethanol is $44.5 \mathrm{mm} \mathrm{Hg}$ and that of methanol is $88.7 \mathrm{mm} \mathrm{Hg}$ at $20°\mathrm{C}$.

Calculate the partial pressure and the total vapour pressure of this solution obtained by mixing $60 \mathrm{g}$ of ethanol with $40 \mathrm{g}$ of methanol, and the mole fraction of ethanol in the vapour.

What are partial and total vapour pressures at $25° \mathrm{C}$ above the solution having equal numbers of molecules of benzene and toluene? The vapour pressures of benzene and toluene at this temperature are $95.1 \mathrm{mm}$ and $28.4 \mathrm{mmHg}$, respectively. What is the composition of the vapour in equilibrium with the benzene-toluene solution?