A) \[224\,c{{m}^{3}}\]
B) 1.008 g
C) \[112\,c{{m}^{3}}\]
D) \[22400\,c{{m}^{3}}\]
Correct Answer: C
Solution :
According to Faradays second law of electrolysis, we have \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\] \[\therefore \] \[\frac{1.08}{x}=\frac{108}{1}\] \[\therefore \] weight of hydrogen \[(x)=0.01\,g\] Hence, the volume of hydrogen at \[STP=\frac{22400\times 0.01}{2}=112\,\,c{{m}^{3}}\]You need to login to perform this action.
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