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

Options

A. 2.7 g
B. 2.7 mg
C. 0.27 g
D. 0.54 g

Answer

C

Explanation

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