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

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