Question

Calculate vapour pressure of a mixture containing $252g$ of $n-$pentane (MW=72) and $1400g$ of $n-$heptane (MW=100) at ${20}^{o}C$. The vapour pressure of $n-$pentane $=420$ mm of $Hg$ and $n-$heptane $=36$ mm of $Hg$.

Answer 1

Solution:-

According to Raoult's law, the vapour pressure exercised by a component of a mixture can be calculated as follows

$P=P_{\circ }x$

whereas,

$P$ is the vapour pressure of the component in the mixture.

$P_{\circ }$ is the vapour pressure of the pure component.

$x$ is the molar fraction of the component in the mixture.

As we know that,

No. of moles $=\frac{\text{Wt.}}{\text{Mol. wt.}}$

Given:-

Wt. of $n$-pentane $=252g$

Mol. wt. of $n$-pentane $=72g$

No. of moles of $n$-pentane $=\frac{252}{72}=3.5$

Wt. of $n$-heptane $=1400g$

Mol. wt. of $n$-pentane $=100g$

No. of moles of $n$-pentane $=\frac{1400}{100}=14$

Total no. of moles $=3.5+14=17.5\text{moles}$

Now as we know that,

Mole fraction $=\frac{\text{No. of moles}}{\text{Total no. of moles}}$

Therefore,

$x_{n-pentane}=\frac{3.5}{17.5}=0.2$

$x_{n-heptane}=\frac{14}{17.5}=0.8$

Therefore

$P_{n-pentane}=0.2\times 420=84\text{mm Hg}$

$P_{n-heptane}=0.8\times 36=28.8\text{mm Hg}$

Therefore, the vapour pressure of mixture is-

$P_{mixture}=84+28.8=112.8\text{mm Hg}$

Hence the vapour presure of mixture is $112.8\text{mm Hg}$.