A) 64
B) 32
C) 16
D) 8
E) 4
Correct Answer: A
Solution :
Relation between rate of diffusions and molar masses of the gases X and\[C{{H}_{4}}\]is \[\frac{{{r}_{x}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{{{M}_{C{{H}_{4}}}}}{{{M}_{x}}}}\] Let the rate of diffusion of gas \[X=a\] \[\therefore \] \[\frac{a}{2a}=\sqrt{\frac{16}{{{M}_{X}}}}\] (\[\because \]\[C{{H}_{4}}=12.4=16\]) \[\therefore \] \[\frac{1}{2}=\sqrt{\frac{16}{{{M}_{X}}}}\] \[\therefore \] \[{{M}_{X}}=16\times 4=64\] So, the molar mass of gas X is 64.
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