A) 1:2
B) 2:1
C) 1 :1
D) 1:4
Correct Answer: B
Solution :
According to Grahams law of diffusion, \[\frac{{{r}_{S{{O}_{2}}}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{{{M}_{C{{H}_{4}}}}}{{{M}_{S{{O}_{2}}}}}}\] \[\frac{{{r}_{S{{O}_{2}}}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{16}{64}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\] \[{{r}_{S{{O}_{2}}}}:{{r}_{C{{H}_{4}}}}=1:2\] \[\text{S}{{\text{O}}_{\text{2}}}\] and \[\text{C}{{\text{H}}_{4}}\] effused in the ratio 1 : 2. So, the amount remains in the vessel will be in the ratio as \[S{{O}_{2}}:C{{H}_{4}}=2:1\]You need to login to perform this action.
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