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

Answer:

73.43kPa.

Explanation:

We know that the molar mass of the octane (C8H18) is 114g/mol and that of the heptane (C7H16) is 100g/mol.

So, number of moles of the heptane=26/100=0.26mol

The number of moles of octane is 35/114 = 0.31mol.

So, the Xa of the heptane is 0.26/0.26+0.31=0.456.

And of that of the octane is = 0.31/0.26+0.31=0.544 .

Hence, the vapour pressure of the heptane=0.456×105.2 kPa  = 47.97 kPa.

And that of the octane will be =0.544×46.8 kPa  = 25.46kPa.

So, the total vapour pressure of the mixture will be Ptotal = 47.97+25.46  73.43kPa.

Answer 2

Answer:

Total vapour pressure of the mixture = 48.18+25.36=73.54 KPa

Explanation:

The molar mass of Heptane ($C_{7}H_{16}$) = 100 g/mol

The molar mass of Octane ($C_{7}H_{16}$) =114 g/mol

No. of mole of Heptane ($C_{7}H_{16}$) = $\dfrac{26}{100} =0.26$

No. of mole of Octane  ($C_{7}H_{16}$) = $\dfrac{35}{114} =0.307$

Mole fraction of Heptane ($C_{7}H_{16}$) =$\dfrac{No.\ of\ moles\of\ heptane}{No.\ of\ moles\of\ heptane+No.\ of\ moles\of\ octane} = \dfrac{0.26}{0.26+0.307} =0.458$

Mole fraction of Octane ($C_{7}H_{16}$) =1-Mole fraction of heptane=1-0.458=0.542

Vapour pressure of Heptane =$0.458\times 105.2=48.18\ KPa$

Vapour pressure of Octane = $0.542\times 46.8=25.36\ KPa$

Total vapour pressure of the mixture = 48.18+25.36=73.54 KPa