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States Of Matter

Question
CBSEENCH11005933

2.9 g of a gas at 95°C occupied the same volume as 0·184g of hydrogen at 17°C at the same pressure. What is the molar mass of the gas? 

Solution

For gas:
                PV space equals space nRT               ...(1)
Here comma space space straight n space equals space fraction numerator Mass space of space gas over denominator Molecular space mass space of space gas end fraction space equals fraction numerator 2.9 space straight g over denominator straight M subscript gas end fraction
space space space space space space space space space straight T space equals space 95 plus 273 space equals space 368 space straight K
Substituting the values  in eq. (1), we have,
               PV space equals space fraction numerator 2.9 straight g over denominator straight M subscript gas end fraction cross times straight R cross times 368 straight K space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
For hydrogen gas:
  straight n space equals space fraction numerator Mass space of space hydrogen over denominator Molecular space mass space of space hydrogen end fraction
space space space space equals space fraction numerator 0.184 space straight g over denominator 2 straight g space mol to the power of negative 1 end exponent end fraction
straight T space equals space 17 plus 273 space equals space 290 space straight K
therefore Substituting the values in eq. (1) we have,
   
 PV space equals space fraction numerator 0.184 space straight g over denominator 2 straight g space mol to the power of negative 1 end exponent end fraction cross times space straight R space cross times space 290 straight K   ...(3)
.
From eq. (2) and (3), we have
fraction numerator 2.9 straight g over denominator straight M subscript gas end fraction cross times space straight R cross times 368 space straight K space equals space fraction numerator 0.184 space straight g over denominator 2 straight g space mol to the power of negative 1 end exponent end fraction space cross times space straight R space cross times space 290 space straight K
therefore space space space space straight M subscript gas space equals space fraction numerator 2.9 space cross times space 368 space cross times space 2 over denominator 0.184 space cross times space 280 end fraction straight g space mol to the power of negative 1 end exponent
space space space space space space space space space space space space space space equals space fraction numerator 2134.4 space over denominator 53.36 end fraction straight g space mol to the power of negative 1 end exponent space equals space 40 space straight g space mol to the power of negative 1 end exponent