`3.7 g` of a gas at `25^(@)C` occupied the same volume as `0.184 g` of hydrogen at `17^(@)C` and at the same pressure. The molecular mass of the gas is

Options

A. 0.024
B. 39.14
C. 41.33
D. 59.14

Answer

C

Explanation

For hydrogen, w = 0.184 g , T = 17 + 273 = 290 K , M = 2
From pV = `(w)/(M) RT = (0.184)/(2) xx R xx 290 " "` … (i)
For unknown gas , w = 3.7 g T = 25 + 273 = 298 K , M = ?
`pV = (3.7)/(M) xx R xx 298`
From Eqs. (i) and (ii) `(3.7)/(M) xx R xx 298 = (0.184)/(2) xx R xx 290 or M = (3.7 xx 298 xx 2)/(0.184 xx 290) = 41.33`