A current strength of 3.86 amp was passed through molten calcium oxide for 41 minutes and 40 seconds. The mass of calcium in grams deposited at the cathode is (Atomic mass of Ca is 40 g/mol, 1 F = 96500 C)

Question

A current strength of 3.86 amp was passed through molten calcium oxide for 41 minutes and 40 seconds. The mass of calcium in grams deposited at the cathode is (Atomic mass of Ca is 40 g/mol, 1 F = 96500 C) (A) 4 (B) 2 (C) 6 (D) 8

Tardigrade Admin · Accepted Answer

I=3.86 Amp t=41 min 40 sec =2500 seconds Mass deposited =Z I t [Z=( text Equivalent weight /96500)] Equivalent wt. of C a=(40/2)=20 Hence mass deposited =(20/96500) × 2500 × 3.86=2 g

Q. A current strength of 3.86 amp was passed through molten calcium oxide for 41 minutes and 40 seconds. The mass of calcium in grams deposited at the cathode is (Atomic mass of Ca is 40 g/mol, 1 F = 96500 C)

4

2

6

8

1

$I = 3.86$ Amp
$t = 41 min 40 sec = 2500$ seconds
Mass deposited $= Z I t$
$[Z = \frac{Equivalent weight}{96500}]$
Equivalent wt. of $C a = \frac{40}{2} = 20$
Hence mass deposited
$= \frac{20}{96500} \times 2500 \times 3.86 = 2 g$

Q. A current strength of 3.86 amp was passed through molten calcium oxide for 41 minutes and 40 seconds. The mass of calcium in grams deposited at the cathode is (Atomic mass of Ca is 40 g/mol, 1 F = 96500 C)

4

2

6

8

1

I=3.86 Amp t=41min40sec=2500 seconds Mass deposited =ZIt [Z=96500 Equivalent weight ​] Equivalent wt. of Ca=240​=20 Hence mass deposited =9650020​×2500×3.86=2g

$I = 3.86$ Amp
$t = 41 min 40 sec = 2500$ seconds
Mass deposited $= Z I t$
$[Z = \frac{Equivalent weight}{96500}]$
Equivalent wt. of $C a = \frac{40}{2} = 20$
Hence mass deposited
$= \frac{20}{96500} \times 2500 \times 3.86 = 2 g$