# Heptane and octane form an ideal solution. At 373 K,

Question:

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 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?

Solution:

Vapour pressure of heptane $\left(p_{1}^{0}\right)=105.2 \mathrm{kPa}$

Vapour pressure of octane $\left(p_{2}^{0}\right)=46.8 \mathrm{kPa}$

We know that,

Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1

= 100 g mol−1

$\therefore$ Number of moles of heptane $=\frac{26}{100} \mathrm{~mol}$

= 0.26 mol

Molar mass of octane (C8H18) = 8 × 12 + 18 × 1

= 114 g mol−1

$\therefore$ Number of moles of octane $=\frac{35}{114} \mathrm{~mol}$

= 0.31 mol

Mole fraction of heptane, $x_{1}=\frac{0.26}{0.26+0.31}$

= 0.456

And, mole fraction of octane, x2 = 1 − 0.456

= 0.544

Now, partial pressure of heptane, $p_{1}=x_{1} p_{1}^{0}$

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane, $p_{2}=x_{2} p_{2}^{0}$

= 0.544 × 46.8

= 25.46 kPa

Hence, vapour pressure of solution, ptotal p1 + p2

= 47.97 + 25.46

= 73.43 kPa