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?

Answer 1

The correct answer is: $= 73.43 kPa$
Vapour pressure of heptane$(p^0_1)= 105.2KPa$
Vapour pressure of octane$= 46.8 kPa$
We know that,
Molar mass of heptane $(C_7H_{16}) = 7\times12 + 16 \times 1$
$= 100 g mol ^{- 1}$
Number of moles of heptane $= \frac{26}{100} mol$
$= 0.26 mol$
Molar mass of octane $(C_8H_{18}) = 8\times12 + 18 \times 1$
$= 114 g mol^{- 1}$
Number of moles of octane$=\frac{35}{114} mol$
$= 0.31 mol$
Mole fraction of heptane, $x_1 = \frac{0.26}{(0.26+0.31)}$
$= 0.456$
And, mole fraction of octane, $x_2 = 1 - 0.456$
$= 0.544$
Now, partial pressure of heptane, $p_1 = x_1p^0_1$
$= 0.456\times105.2 = 47.97 kPa$
Partial pressure of octane, $p_2 = x_2p^0_2$
$= 0.544\times46.8 = 25.46 kPa$
Hence, vapour pressure of solution, $p_{total} = p_1 + p_2$
$= 47.97 + 25.46$
$= 73.43 kPa$