NCERT Class XII Chemistry
Chapter - Solutions
Questions with Solutions
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Question : 38
Total: 53
Benzene and naphthalene form ideal solution over the entire range of composition. The vapour pressures of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.
Solution:
Molar mass of benzene ( C 6 H 6 ) = 78 g mol – 1
Molar mass of naphthalene( C 10 H 8 ) = 128 g mol – 1
∴ Number of moles in 80 g of benzene =
= 1.026 mole
∴ Number of moles in 100 g of naphthalene=
∴ In the solution, mole fraction of benzene=
=
Mole fraction of naphthalene = 1 – 0.567= 0.433
P 0 Benzene = 50.71 P 0 naphthalene = 32.06
Applying Raoult’s law of vapour pressure
P Benzene = x Benzene × P 0 Benzene
P naphthalene = x naphthalene × P 0 naphthalene
∴ From Dalton’s law, mole fraction of benzene in the vapour phase
=
=
=
= 0.6744
Molar mass of naphthalene
∴ Number of moles in 80 g of benzene =
∴ Number of moles in 100 g of naphthalene
∴ In the solution, mole fraction of benzene
Mole fraction of naphthalene = 1 – 0.567= 0.433
Applying Raoult’s law of vapour pressure
∴ From Dalton’s law, mole fraction of benzene in the vapour phase
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