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How long a current of 3 amperes has to be passed through a solution of silver nitrate to coat a metal surface of 80 cm^2 with 0.005 mm thick layer ? [Density of Ag is 10.5 g cm^(-3), At wt.of silver = 108.0 u] |
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Answer» Solution :`Ag^(+) (aq) + E^(-) to Ag (s)` Volume = Area x Thickness `=80 cm^(2) XX 0.05 xx 1/10 cm = 0.04 cm^(3)` Mass of Ag = Volume x Density `=0.04 cm^(3) xx 10.5 G cm^(-3) = 0.42 g` Let the time required be t seconds. Total charge passed - 3T COULOMBS. 96500 coulombs of charge deposits = 108 g 3t coulombs of charge deposits `=108/96500 xx 3t g` Thus, `108/96500 xx 3t = 0.42` or `t = (0.42 xx 96500)/(108 xx 3) = 125.09 s` |
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