Correct Answer - C
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