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Fakaruddin Raj Nagi
Asked: 2022-11-02T23:26:09+05:30 In:

`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)`

`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)`
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  1. Correct Answer – 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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