The no. of coulombs required to liberate 0.224 dm^(3) of chlorine at 0 – InterviewSolution

1.	The no. of coulombs required to liberate 0.224 dm^(3) of chlorine at 0^(@)C and 1 atm passed are :
Answer» `2 XX 965` 96500 96.5 965/2 Solution :`22.4 dm^3 of Cl_2=1" MOLE of "Cl_2=2F` `THEREFORE 22.4 dm^3 =2F` Then 0.224 =0.02F `therefore` Coulombs =`0.02xx96500 or 2xx965`

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