A) 0.500 g
B) 8.00 g
C) 4.00 g
D) 16.00 g
Correct Answer: C
Solution :
According to Faraday?s I law,\[\omega =Z\] it Given, the value of it is same for \[{{\operatorname{H}}_{2}} and {{O}_{2}}\]. Then, \[\frac{{{\omega }_{{{H}_{2}}}}}{{{\omega }_{{{O}_{2}}}}}=\frac{{{Z}_{{{H}_{2}}}}}{{{Z}_{{{O}_{2}}}}}\] \[\frac{{{\omega }_{{{H}_{2}}}}}{{{\omega }_{{{O}_{2}}}}}=\frac{{{E}_{{{H}_{2}}}}}{{{E}_{{{O}_{2}}}}}\] \[\frac{0.500}{{{\omega }_{{{O}_{2}}}}}=\frac{2}{16}\] \[{{\omega }_{{{O}_{2}}}}=\frac{16\times 0.5}{2}\] \[{{\omega }_{{{O}_{2}}}}=4.00\,\,g\]
You need to login to perform this action.
You will be redirected in
3 sec
