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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Jul 2007 09:08:49 IST
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An electric current of 1 A is passed through acidulated water for 160 minutes and 50 second. What is the volume of the oxygen liberated at the anode(as reduced to NTP)?
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28 Jul 2007 14:08:27 IST
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1 mole of oxygen (22.4 Ltr) requires 4Faraday charge= 4*96500 C
Charge supplied = i*t = 1*(160*60+50)=9650C
So vol of O2 produced = 22.4*9650/(4*96500)=22.4/40 ltr
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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