Calculate the mass of copper deposited if a current of 0.45A flows through (CuSO_{4}) solution for 1hour 15mins. [Cu=64.0, S= 32.0, O =16.0, 1F= 96500C]
-
A.
6.40g -
B.
0.67g -
C.
0.64g -
D.
0.45g
Correct Answer: Option B
Explanation
(Cu^{2+} + 2e^{-} to Cu_{(s)})
( 2 times 96500C = 193000C deposits 64g of Cu)
(M propto It)
(Q = It = 0.45A times 1hr15mins)
= ( 0.45 times 75 times 60) [converting the time to seconds]
Quantity of electricity passed = 2025C
(193000C to 64g Cu)
(1C to frac{64}{193000})
(2025C to frac{64}{193000} times 2025)
(= 0.6715g approxeq 0.67g)