A) 0.5 g
B) 4 g
C) 6 g
D) 8 g
Correct Answer: A
Solution :
\[{{r}_{H}}=\frac{2gm}{10\min }\] if \[{{r}_{O}}=\frac{xgm}{10\min }\] \[{{r}_{O}}={{r}_{H}}\sqrt{\frac{{{M}_{{{H}_{2}}}}}{{{M}_{{{O}_{2}}}}}}=\frac{2}{10}\sqrt{\frac{2}{32}}\] \[\frac{x}{10}=\frac{2}{10\times 4}=\frac{1}{2}gm.=.5gm\]
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