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Question

A current of 9.65A is passed for 3 hours between nickel electrodes immersed in 0.5 dm3 solution of concentration 2 mol dm3 Ni(NO3)2.

The molarity of the solution after electrolysis would be :

A
0.46M
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B
0.92M
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C
1.25M
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D
0.625M
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Solution

The correct option is C 0.92M
The moles of nickel initially present =2mol1dm3×0.5dm3=1mol.
The quantity of electricity passed =Q(C)=9.65A×10800s=104220C.
The moles of electrons passed =Q(C)96500(Cmole)=104220C96500(Cmole)=1.08mole.
The moles of Ni deposited =moleratio×molesofelectronspassed=1molNi2mole×1.08mole=0.54mol.
The moles of nickel remaining =1mol0.54mol=0.46mol.
The molarity of the solution is 0.46mol0.5dm3=0.92M.

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