Question

Equal weights of ethane and hydrogen are mixed in an empty container $25^o$C. The fraction of the total pressure exerted by hydrogen is
A.1:2
B.1:1
C.1:16
D.15:16

Answer 1

Hint : We must know that the partial pressure of a gas or component in a mixture is equal to its mole fraction times the total pressure of the gas mixture

Complete step by step solution :
As given in question,
Ethane, ${C_2}{H_5}$, and Hydrogen, ${H_2}$are mixed in an equal proportion of weights.
Consider ‘x’ gm of each of Ethane and Hydrogen are mixed.
∴ Mole of ethane = $\dfrac{x}{{30}}$ (molecular weight of ethane is 30)
∴ Mole of hydrogen = $\dfrac{x}{2}$ (molecular weight of hydrogen is 2)
∴ Mole fraction of hydrogen $= \dfrac{{\dfrac{x}{2}}}{{\dfrac{x}{2} + \dfrac{x}{{30}}}}$
By solving above, we get
∴ Mole fraction of hydrogen $= \dfrac{{15}}{{16}}$
We know that the partial pressure of each component in a mixture is proportional to its mole fraction. Hence, in the mixture of gases, the partial pressure of each component is the product of the total pressure and the mole fraction of that component.
Corresponding to Laws of Partial pressures of components in the mixture, as the partial pressure of the component in a mixture is equal to its mole fraction times the total pressure of the gas mixture
∴ mole fraction of hydrogen =$\dfrac{{{\text{Partial pressure of hydrogen}}}}{{{\text{Total pressure }}}}$ = $\dfrac{{15}}{{16}}$ = 15:16
so, the fraction of the total pressure exerted by hydrogen is 15:16
Hence, We can conclude that the option (D) is the correct option.

Note : We must know that the sum of the mole fractions of all the components in the mixture will always be equal to 1.We must know the relationship between partial pressures and mole fractions of gases or components in a mixture. In a mixture,the partial pressure of a gas in a mixture is equal to its mole fraction times the total pressure of the gas mixture.