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2.9 g of a gas at 95^(@) C occupied the ...

`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 ? .

A

`40 g mol^(-1)`

B

`50 g mol^(-1)`

C

`60 g mol^(-1)`

D

`30 g mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
According to the ideal gas law `(pV = nRT)`, we have
`p_(gas)V_(gas) = n_(gas)RT_(gas)`
`p_(H_(2)) = n_(H_(2))RT_(H_(2))`
Since `p_(gas) = pH_(2)` and `V_(gas) = V_(H_(2))`
`n_(gas)RT_(gas) = n_(H_(2))RT_(H_(2))`
or `n_(gas) T_(gas) = n_(H_(2))T_(H_(2))`
Since `n = (Mass(m))/(Molar mass(M))`
we can write `(m_(gas))/(M_(gas)) T_(gas) = (m_(H_(2)))/(M_(H_(2)))T_(H_(2))`
or `M_(gas) = (M_(H_(2))m_(gas)T_(gas))/(m_(H_(2))T_(H_(2)))`
`= ((2 g mol^(-1))(2.9g) (95 + 273 K))/((0.184 g)(17 + 273 K))`
`= 40 g mol^(-1)`
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