A current of electricity is passed through a copper voltameter
and a water voltameter connected in series if the copper anode of the copper voltameter now weighs 16 mg less ,hydrogen liberated at the cathode of the water voltameter measures at stp about
4.0ml
5.6ml
6.4ml
8.0ml
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1 Answers
1357mmol of Cu lost, n=16/63.5
meq=16*2/63.5
meq of H2=16*2/63.5
mmol if H2=16/63.5
volume=(16/63.5)*22.4 ml≈5.6 ml