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NTA Abhyas NEET2020ChemistryElectrochemistryPractice

The amount of silver deposited by passing 241.25 C of charge through silver nitrate solution is

Options

  1. A2.7 g
  2. B2.7 mg
  3. C0.27 g
  4. D0.54 g

Correct answer

C. 0.27 g

Step-by-step solution

Given, current = 241.25 C We know that 1 C electricity will deposit 1.118 × 10 - 3 g of silver. ∴ 241.25 C electricity will deposit = ( 1.118 × 10 - 3 ) × 241.25 = 0.27 g of silver.

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