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Question: A steady current of 5 amps is maintained for 45 mins. During this time it deposits 4.572 gms of zinc...

A steady current of 5 amps is maintained for 45 mins. During this time it deposits 4.572 gms of zinc at the cathode of a voltameter. E.C.E. of zinc is

A

3.387×10−4gm/C3.387 \times 10^{- 4}gm/C

B

3.387×10−4C/gm3.387 \times 10^{- 4}C/gm

C

3.384×10−3gm/C3.384 \times 10^{- 3}gm/C

D

3.394×10−3C/gm3.394 \times 10^{- 3}C/gm

Answer

3.387×10−4gm/C3.387 \times 10^{- 4}gm/C

Explanation

Solution

m=Zitm = Zit ⇒ Z=mit=4.5725×45×60=3.387×10−4gm/CZ = \frac{m}{it} = \frac{4.572}{5 \times 45 \times 60} = 3.387 \times 10^{- 4}gm/C