A) 1:4
B) 4:1
C) 1 : 8
D) 8 : 1
Correct Answer: A
Solution :
According to Grahams rule \[\frac{^{r}{{O}_{_{2}}}}{^{r}{{H}_{2}}}=\sqrt{\frac{{{M}_{{{H}_{2}}}}}{{{M}_{{{O}_{2}}}}}}\] \[=\sqrt{\frac{2}{32}}\] \[=\frac{1}{4}\] \[\therefore \] \[^{r}{{O}_{2}}{{:}^{r}}{{H}_{2}}=1:4\]You need to login to perform this action.
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