A current of 4 A is passed through a solution of silver nitrate to coat a metal surface of 80 `cm^(2)` with 0.005 mm thick layers of silver. If the density of silver (molar mass: 108 g `"mol"^(-1)`)is 10.8 g `cm^(-3)` , the time for which the current is passed is about

Options

A. 84.5s
B. 90.5s
C. 96.5s
D. 100.5s

Answer

C

Explanation

Volume of Ag deposited `=(80 cm^(2)) (5 xx 10^(-1)cm)= 0.04cm^(3)`
Mass of Ag deposited `=rho V= (10.8 g cm^(-3)) (0.04 g cm^(3)) = 0.432g`
Amount of Ag deposited `=(0.432g)/(108 g mol^(-1))= 0.004` mol
Quantity of electricity passed = (0.004 mol) (96500C `mol^(-1)`) = 386C
Time for which 4A current is passed `=(386C)/(4A)= 96.5s`