A current of 1.40 ampere is passed through 500 mL of 0.180 M solution of zinc sulphate for 200 seconds. What will be the molarity of Zn2+ ions after deposition of zinc?

Question

A current of 1.40 ampere is passed through 500 mL of 0.180 M solution of zinc sulphate for 200 seconds. What will be the molarity of Zn2+ ions after deposition of zinc? (A) 0.154 M (B) 0.177 M (C) 2 M (D) 0.180 M

Tardigrade Admin · Accepted Answer

Amount of charge passed = 1.40 × 200 = 280 Coulombs No. of moles of Zn deposited by passing 280 C of charge =(1/2×96500)×280=0.00145 Molarity of zinc after deposition of zinc =0.180-(0.00145×1000/500) =0.180-0.0029 =0.177 M

Q. A current of $1.40$ ampere is passed through $500 m L$ of $0.180 M$ solution of zinc sulphate for $200$ seconds. What will be the molarity of $Z n^{2 +}$ ions after deposition of zinc?

$0.154 M$

$0.177 M$

$2 M$

$0.180 M$

Amount of charge passed $= 1.40 \times 200$
$= 280$ Coulombs
No. of moles of $Z n$ deposited by passing $280 C$ of charge
$= \frac{1}{2 \times 96500} \times 280 = 0.00145$
Molarity of zinc after deposition of zinc
$= 0.180 - \frac{0.00145 \times 1000}{500}$
$= 0.180 - 0.0029$
$= 0.177 M$

Q. A current of 1.40 ampere is passed through 500mL of 0.180M solution of zinc sulphate for 200 seconds. What will be the molarity of Zn2+ ions after deposition of zinc?

0.154M

0.177M

2M

0.180M