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

From the available given
No. of moles of gas `= ("Mass of gas")/("Molar mass") = ((2.9 g))/((M))`
No. of mass of hydrogen `= ("Mass of hydrogen")/("Molar mass") = ((0.184 g))/((2g mol^(-1))) = 0.092` mol
Temperature of the gas `(T_(1)) = 95 + 273 = 368 K`
Temperature of hydrogen `(T_(2)) = 17 + 273 = 290 K`
According to ideal has equation `PV = nRT`
For the gases, P, V and R are constant.
`:. n_((g)) xx T_(1) = n_(H_(2)) xx T_(2)` or `n_((g)) = (n_(H_(2)) xx T_(2))/(T_(1))`
`((2.9 g))/((M)) = ((0.092 mol) xx (290 K))/((368 K))`
or `M = ((2.9 g) xx (368 K))/((0.092 mol) xx (290 K)) = 40 g mol^(-1)`.