Question

QUESTION 2.156

Heptane and octane form an ideal solution. At 373 K, the vapour pressure of the two liquid components is 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Answer 1

Number of moles of octane $\left(n_A\right)=\frac{Mass}{Molarmass}=\frac{35g}{114gmo{l}^{-1}}$
=0.307 mol
[Molar mass of octane
$C_8{H}_{18}=(12\times 8)+(1\times 18)=114gmo{l}^{-1}$]
Number of moles of heptane
$\left(n_B\right)=\frac{26g}{100gmo{l}^{-1}}=0.26mol$
[Molar mass of heptane
$C_7{H}_{16}=(12\times 7)+16=100gmo{l}^{-1}$]
Mole fraction of octane
$\left(x_A\right)=\frac{n_A}{n_A+n_B}=\frac{\left(0.307mol\right)}{(0.307+0.26)mol}=0.542$

Mole fraction of Heptane
$\left(x_B\right)=\frac{n_B}{n_A+n_B}=\frac{\left(0.26mol\right)}{(0.307+0.26)mol}=0.458$
Vapour pressure of pure heptane $\left(p_B^{\circ }\right)=105.2kPa$
Vapour pressure of pure octane $\left(p_A^{\circ }\right)=46.8kPa$
In the mixture of 26.0 g heptane and 35.0 g octane
(i) Vapour pressure of heptane $\left(p_B\right)=p_B^{\circ }{x}_B$
$=(105.2kPa\times 0.458)$
$=48.18kPa$
(ii) Vapour pressure of octane
$\left(p_A\right)=p_A^{\circ }{x}_A=(46.8kPa\times 0.542)$
$=25.36kPa$
(iii) Total vapour pressure of the mixture (p) $=p_A+p_B$
= 25.36 + 48.18 = 73.54 kPa