`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 ? .

Answer

`40gmol^(-1)`

Explanation

Volume (V) occupied by dihydrogen is given by,
`V=m/M(RT)/P`
`=(0.184)/2xx(Rxx290)/p`
Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as:
`V=m/M(RT)/P`
`=(2.9)/(M)xx(Rxx368)/(p)`
According to the questin,
`(0.184)/(2)xx(Rxx290)/(p)=(2.9)/(M)xx(Rxx368)/(p)`
`rArr(0.184xx290)/(2)=(2.9xx368)/(M)`
`rArrM=(2.9xx368xx2)/(0.1184xx290)`
= 40 g `mol^(-1)`
Hence, the molar mass of the gas is 40 g `mol^(-1).`