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  4. The amount of silver deposited by passing 241.25 coulomb of current through silver nitrate solution is

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Q. The amount of silver deposited by passing 241.25 coulomb of current through silver nitrate solution is

VMMC MedicalVMMC Medical 2004

Solution:

Given : current = 241.25 coulomb We know that 1 coulomb of electricity will deposit $ 1.118\times {{10}^{-3}}\,g $ of silver. $ \therefore $ 241.25 coulomb electricity will deposit $ =(1.118\times {{10}^{-3}})\times 241.25 $ = 0.27 g of silver.