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

Question

The amount of silver deposited by passing 241.25 coulomb of current through silver nitrate solution is (A) 2.7g (B) 2.7mg (C) 0.27g (D) 0.54g

Tardigrade Admin · Accepted Answer

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

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

2.7g

2.7mg

0.27g

0.54g

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

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

2.7g

2.7mg

0.27g

0.54g

Given : current = 241.25 coulomb We know that 1 coulomb of electricity will deposit 1.118×10−3g of silver. ∴ 241.25 coulomb electricity will deposit =(1.118×10−3)×241.25 = 0.27 g of silver.

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