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

Explanation

As `P_(1) = P_(2) and v_(1) = v_(2)`, But PV = nRT
`:. P_(1) V_(1) = P_(2) V_(2)`, i.e., `n_(1) RT_(1) = n_(2)RT_(2) " " :. n_(1) T_(1) = n_(2) T_(2)` or `(w_(1))/(M_(1)) T_(1) = (w_(2))/(M_(2)) T_(2)`
i.e., `(2.9)/(M_(x)) xx (95 + 273) = (0.184)/(2) xx (17 + 273) or M_(x) = (2.9 xx 368 xx 2)/(0.184 xx 290) = 40 g mol^(-1)`