A current of 3.7 ampere is passed for 6 hr between Ni electrodes in 0.5 litre of 2 M solution of Ni(NO3)2 . What will be the molarity of solution at the end of electrolysis?

Time=6hr=6 x 60 x 60=21600m=E x3.796500x21600=592(96500)21600x3.7M=6.6 x 3.7=24.42M=WtGMWx10.5M=24.425.9x10.5=0.890.89M<2M

A current of 3.7 ampere is passed for 6 hr between Ni electrodes in 0.5 litre of 2 M solution of Ni(NO3)2 . What will be the molarity of solution at the end of electrolysis?

Time=6hr=6 x 60 x 60=21600m=E x3.796500x21600=592(96500)21600x3.7M=6.6 x 3.7=24.42M=WtGMWx10.5M=24.425.9x10.5=0.890.89M<2M

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A current of 3.7 ampere is passed for 6 hr between Ni electrodes in 0.5 litre of 2 M solution of Ni(NO₃)₂ . What will be the molarity of solution at the end of electrolysis?

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$Time = 6 hr = 6 x 60 x 60 = 21600$

$m = \frac{E x 3.7}{96500} x 21600 = \frac{59}{2 (96500)} 21600 x 3.7$

$M = 6.6 x 3.7 = 24.42$

$M = \frac{Wt}{GMW} x \frac{1}{0.5}$

$M = \frac{24.42}{5.9} x \frac{1}{0.5} = 0.89$

$0.89 M < 2 M$

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A current of 3.7 ampere is passed for 6 hr between Ni electrodes in 0.5 litre of 2 M solution of Ni(NO3)2 . What will be the molarity of solution at the end of electrolysis?

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A current of 3.7 ampere is passed for 6 hr between Ni electrodes in 0.5 litre of 2 M solution of Ni(NO₃)₂ . What will be the molarity of solution at the end of electrolysis?