When 3.86 A current is passed through an electrolyte for 50 min, 2.4 g of a divalent metal is deposited. The gram atomic weight of the metal (in grams) is

Question

When 3.86 A current is passed through an electrolyte for 50 min, 2.4 g of a divalent metal is deposited. The gram atomic weight of the metal (in grams) is (A) 24 (B) 12 (C) 64 (D) 40

Tardigrade Admin · Accepted Answer

w=Zit or w=(Eq.weight/96500)× i× t or w=(Atomic weight/Valency× 96500)× i× t Atomic weight =(w× valency× 96500/i× t) =(2.4× 2× 96500/3.86× 50× 60)=40

Q. When 3.86 A current is passed through an electrolyte for 50 min, 2.4 g of a divalent metal is deposited. The gram atomic weight of the metal (in grams) is

24

12

64

40

$w = Z i t$ or $w = \frac{Eq . w e i g h t}{96500} \times i \times t$ or $w = \frac{A t o mi c w e i g h t}{Va l e n cy \times 96500} \times i \times t$ Atomic weight $= \frac{w \times v a l e n cy \times 96500}{i \times t}$ $= \frac{2.4 \times 2 \times 96500}{3.86 \times 50 \times 60} = 40$

Q. When 3.86 A current is passed through an electrolyte for 50 min, 2.4 g of a divalent metal is deposited. The gram atomic weight of the metal (in grams) is

24

12

64

40

w=Zit or w=96500Eq.weight​×i×t or w=Valency×96500Atomicweight​×i×t Atomic weight =i×tw×valency×96500​ =3.86×50×602.4×2×96500​=40

$w = Z i t$ or $w = \frac{Eq . w e i g h t}{96500} \times i \times t$ or $w = \frac{A t o mi c w e i g h t}{Va l e n cy \times 96500} \times i \times t$ Atomic weight $= \frac{w \times v a l e n cy \times 96500}{i \times t}$ $= \frac{2.4 \times 2 \times 96500}{3.86 \times 50 \times 60} = 40$