Question

At $300\mathrm{K}$ the vapour pressure of a solution containing $1$ mole of n-hexane and $3$ moles of n-heptane is $550$ mm of $\mathrm{Hg}$. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of this solution increases by $10$ mm of $\mathrm{Hg}$. What is the vapour pressure in $\mathrm{mm}$ of $\mathrm{Hg}$ of n-heptane in its pure state?

Answer 1

Step 1: Formula for the vapour pressure of a solution containing two components is -

${\mathrm{P}}_{\mathrm{T}}={\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^0{\mathrm{x}}_{\mathrm{B}}$, where ${\mathrm{P}}_{\mathrm{T}}$ is the vapour pressure of the solution, ${\mathrm{P}}_{\mathrm{A}}^0$ and ${\mathrm{P}}_{\mathrm{B}}^0$ are vapour pressure in the pure state, of both the components present in the solution. ${\mathrm{x}}_{\mathrm{A}}$ and ${\mathrm{x}}_{\mathrm{B}}$ are mole fractions of both the components.

Step 2: Vapour pressure of the solution containing $1$ mole of n-hexane and $3$ moles of n-heptane -

${\mathrm{P}}_{\mathrm{T}}=550\mathrm{mm}\mathrm{of}\mathrm{Hg}$

${\mathrm{x}}_{\mathrm{A}}=\frac{1}{4}$

${\mathrm{x}}_{\mathrm{B}}=\frac{3}{4}$

$550=\frac{1}{4}{\mathrm{P}}_{\mathrm{n}-\mathrm{hexane}}^0+\frac{3}{4}{\mathrm{P}}_{\mathrm{n}-\mathrm{heptane}}^0$

$2200={\mathrm{P}}_{\mathrm{n}-\mathrm{hexane}}^0+3{\mathrm{P}}_{\mathrm{n}-\mathrm{heptane}}^0\left(1\right)$

Step 3: Vapour pressure of the solution containing $1$ mole of n-hexane and $5$ moles of n-heptane -

${\mathrm{P}}_{\mathrm{T}}=560\mathrm{mm}\mathrm{of}\mathrm{Hg}$

${\mathrm{x}}_{\mathrm{A}}=\frac{1}{5}$

${\mathrm{x}}_{\mathrm{B}}=\frac{4}{5}$

$560=\frac{1}{5}{\mathrm{P}}_{\mathrm{n}-\mathrm{hexane}}^0+\frac{4}{5}{\mathrm{P}}_{\mathrm{n}-\mathrm{heptane}}^0$

$2800={\mathrm{P}}_{\mathrm{n}-\mathrm{hexane}}^0+4{\mathrm{P}}_{\mathrm{n}-\mathrm{heptane}}^0\left(2\right)$

Step 4; Calculation of ${\mathrm{P}}_{\mathrm{n}-\mathrm{heptane}}^0$ -

On, solving equations $\left(1\right)$ and $\left(2\right)$, we get -

${\mathrm{P}}_{\mathrm{n}-\mathrm{heptane}}^0=600\mathrm{mm}\mathrm{of}\mathrm{Hg}$

${\mathrm{P}}_{\mathrm{n}-\mathrm{hexane}}^0=400\mathrm{mm}\mathrm{of}\mathrm{Hg}$

So, the vapour pressure of n-heptane in its pure state $=600\mathrm{mm}\mathrm{of}\mathrm{Hg}$