`2.9 g` of a gas at `95^(@) C` occupied the same volume as `0.184 g` of hydrogen at `17^(@)C` at same pressure What is the molar mass of the gas ? .

Options

A. `120 g mol^(-1)`
B. `20 g mol^(-1)`
C. `80 g mol^(-1)`
D. `40 g mol^(-1)`

Answer

D

Explanation

For gas `V_("gas") = (nRT)/(P) " " … (i)`
For `H_(2) = V_(H_(2)) = (n_(H_(2)) RT_(H_(2)))/(P_(H_(2))) " " … (ii)`
from eq(i) and (ii)
`(2.9 xx 368)/(M_("gas")) = (0.184 xx 290)/(2)` (since volume and pressure of both gases are same) `M_("gas") = 40 g mol^(-1)`