A current of 9.65A is passed for 3 hours between nickel electrodes immersed in 0.5 dm3 solution of concentration 2 mol dm−3Ni(NO3)2.
The molarity of the solution after electrolysis would be :
A
0.46M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.92M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1.25M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.625M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C0.92M 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 =1mol−0.54mol=0.46mol. The molarity of the solution is 0.46mol0.5dm3=0.92M.