A) 1/8
B) ¼
C) 3/8
D) ½
Correct Answer: A
Solution :
\[{{n}_{{{H}_{2}}}}={{n}_{{{O}_{2}}}}\]and \[{{t}_{{{H}_{2}}}}={{t}_{{{O}_{2}}}}\] According to Graham's law \[\frac{{{r}_{{{H}_{2}}}}}{{{r}_{{{O}_{2}}}}}=\sqrt{\frac{{{M}_{{{O}_{2}}}}}{{{M}_{{{H}_{2}}}}}}\Rightarrow \frac{{{v}_{1}}/{{t}_{1}}}{{{v}_{2}}/{{t}_{2}}}=\sqrt{\frac{32}{2}}\] \[\frac{1/2}{1/x}=\sqrt{16}=4\] \[\frac{x}{2}=4\] \[\therefore \] \[x=8\] \[\therefore \] Fraction of \[{{O}_{2}}=1/8\]You need to login to perform this action.
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