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Question: A current of \(1.40\) ampere is passed through 500 mL of \(0.180\)M solution of zinc sulphate for 20...

A current of 1.401.40 ampere is passed through 500 mL of 0.1800.180M solution of zinc sulphate for 200 seconds. What will be the molarity of Zn2+Zn^{2 +}ions after deposition of zinc?

A

0.154M0.154M

B

0.177M0.177M

C

2M

D

0.180M0.180M

Answer

0.177M0.177M

Explanation

Solution

Amount of charge passed

=1.40×200=280Coulombs= 1.40 \times 200 = 280Coulombs

No. of moles of Zn deposited by passing 280 C of

Charge

=12×96500×280=0.00145= \frac{1}{2 \times 96500} \times 280 = 0.00145

Molarity of zinc after deposition of zinc

=0.180−0.00145×1000500=0.180−0.0029=0.177M= 0.180 - \frac{0.00145 \times 1000}{500} = 0.180 - 0.0029 = 0.177M