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Question: E.C.E. of Cu and Ag are \(7 \times 10^{- 6}\) and \(1.2 \times 10^{- 6}.\) A certain current deposit...

E.C.E. of Cu and Ag are 7×10−67 \times 10^{- 6} and 1.2×10−6.1.2 \times 10^{- 6}. A certain current deposits 14 gm of Cu. Amount of Ag deposited is

A

1.2 gm

B

1.6 gm

C

2.4 gm

D

1.8 gm

Answer

2.4 gm

Explanation

Solution

m1m2=Z1Z2\frac{m_{1}}{m_{2}} = \frac{Z_{1}}{Z_{2}} ⇒ m2=m1Z2Z1=14×1.2×10−67×10−6m_{2} = \frac{m_{1}Z_{2}}{Z_{1}} = \frac{14 \times 1.2 \times 10^{- 6}}{7 \times 10^{- 6}}= 2.4 g