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Question

Find the mass of silver deposited when a current of 1.5 ampere is passed through silver nitrate solution for 5 min.
(Atomic mass of $$Ag$$ is $$108$$ g)


Solution

Molar mass of $$Ag$$ $$=108$$ g mol$$^{-1}$$
1 Faraday $$=96500$$ C
$$Q=i \times t=1.5$$ $$\times 5\times 60=450$$ Coulombs
Weight of the substance deposited $$=ZQ$$
$$Z=\cfrac{M}{nF}$$    n-factor of $$AgNO_{3}$$=1
$$Z=\cfrac{108}{1 \times 96500}$$
Weight of the substance=$$\cfrac{108}{96500} \times 450=\cfrac{54 \times 90}{9650}=0.5$$ g

Physics

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