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 (`p_1^0`) = 105.2 kPa
Vapour pressure of heptane (`p_2^0`) = 46.8 kPa
We know that
Molar mass of heptane (C7H16) = 7 x 12 + 16 x 1
= 100 g mol-1
∴Number of moles of heptane = `26/100` mol
= 0.26 mol
Molar mass of octane (C8H18) = 8 × 12 + 18 × 1
= 114 g mol−1
∴ Number of moles of octane = 35/114 mol
= 0.31 mol
Mole fraction of heptane, `x_1 = 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_1p_1^0`
= 0.456 × 105.2
= 47.97 kPa
Partial pressure of octane `p_2 = x_2p_2^0`
= 0.544 × 46.8
= 25.46 kPa
Hence, vapour pressure of solution, ptotal = p1 + p2
= 47.97 + 25.46
= 73.43 kPa
