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JAMB Chemistry 2017 Paper
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© africaexams.com
Question : 32
Total: 49
Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol
-1
]
0.300g
0.250g
0.2242g
0.448g
Validate
Solution:
M =
Molar mass × Quantity of Electricity
96500 × no of charge
=
M
m
I
T
96500
n
Copper II Chloride = CuCl
2
CuCl
2
→ Cu
2+
+ 2Cl
2
Mass of compound deposited =
Molar mass × Quantity of Electricity
96500 × no of charge
Q = IT
I = 0.5A
T = 45 × 60
T = 2700s
Q = 0.5 × 2700
= 1350c
Molarmass = 64gmol
-1
no of charge = + 2
Mass =
64
×
1350
96500
×
2
Mass = 0.448g
© africaexams.com
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