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Solutions

Question
CBSEENCH12005732

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, in bar of a mixture of 26.0 g of heptane and 35.0 g of octane? 

Solution

Answer:
We have given,

Molar mass of heptane,
C7H16 = 100 g mol–1
Molar mass of octane,
C8H18 = 114 g mol–1
Moles of heptane

= Wt. of heptaneMol. mass of heptane

=26100=0.26

Similarly, Moles of octane


=35114=0.31

Mole fraction heptane

=nAnA+nB

= 0.260.26+0.31=0.456

Mole fraction of octane

= 0.310.26+0.31=0.543

Partial vapour pressure = Mole fraction x Vap.
Pressure of pure component.
Partial vapour pressure of heptane

= 0.456 x 105.2 = 47.97 kPa

Partial vapour pressure of octane
= 0.543 x 46.8 = 25.4 kPa

Total vapour pressure of solution = 73.08 kPa.