Equal weight of `CH_(4) " and " H_(2)` are mixed in an empty container at `25^(@)C`. The fraction of the total pressure exerted by `H_(2)` is

Options

A. `1//9`
B. `1//2`
C. `8//9`
D. `16//17`

Answer

C

Explanation

Pressure exerted by `H_(2)` = mole fraction of `H_(2)xx` total pressure
Suppose w gram of both `CH_(4) and H_(2)` were taken.
Moles of `H_(2)=(w)/(M.W.)=(w)/(2)`
Moles of `CH_(4)=(w)/(16)`
Mole fraction of `H_(2)=(w//2)/((w)/(2)+(w)/(16))=(8)/(9)`
Pressure exerted by `H_(2)=(8)/(9)xx` total pressure.