Question

The vapour density of a mixture of gas a(molecular mass=40) and gas b(molecular mass 80) is 25.The mole % of gas b in the mixture is

Answer 1

Answer:

25

Explanation:

Vapour density of a mixture of gas = a = (molecular mass=40) = 25

Vapour density of a mixture of gas = b = (molecular mass = 80) = 25

Since, vapour density of the mixture is 25, thus,

Molar mass of mixture = 2×25 =50

Let the mole percentage of B in mixture = x.

Thus, the average mass -

= [( 100-x) × 40 + x × 80) / 100

= 50

= x = 25%

Thus, the mole % of gas b in the mixture is 25.

Answer 2

Explanation:

Vapour density of a mixture of gas = a = (molecular mass=40) = 25

Vapour density of a mixture of gas = b = (molecular mass = 80) = 25

Since, vapour density of the mixture is 25, thus,

Molar mass of mixture = 2×25 =50

Let the mole percentage of B in mixture = x.

Thus, the average mass -

= [( 100-x) × 40 + x × 80) / 100

= 50

= x = 25%

Thus, the mole % of gas b in the mixture is 25.