Question

How many molecules of chlorine should be deposited from molten sodium chloride in one minute by a current of 300 milliamperes?

Answer 1

Final Answer : 5.62 * 10^(19) molecules

Steps:
1) We have reaction,
2Cl(-) -----> Cl2 + 2e-
no. of electrons losed per unit of species, n = 2
Current, I =300mA= 0.3 A
Time till current was passed, t = 60 s
Molar Mass of Cl2 = 35.5*2 = 71 g

2) By Faraday's Law of electrolysis,
Mass of Cl2 (Chlorine) deposited,
W : (M/nF)* (Q)

We have, M = 71 g
n = 2
1 F = 96,500C
Q = Charge passed through it = I *t
= 0.3 * 60 = 18 C

3) Then, no. of molecules of Cl2

= (Mass deposited/ Molar Massof Cl2) * ( 6.022 * 10^(23 )
= (W/M) * 6.022 * 10^(23 )
= (M/nFM) * (Q) * (6.022 * 10^(23 )
$= \frac{1}{2 \times 96500} \times 18 \times 6.022 \times {10}^{23} \\ = 5.62 \times {10}^{19}$




Hence, No. of molecules of Cl2 is 5.62 * 10^(19)

Answer 2

Hey !!

Q = I × t

Here, I = 300 / 1000 = 0.3 A

      t = 60 s

      Q = 0.3 A × 60 s = 18 C

       2Cl⁻ ------> Cl₂ + 2e⁻

            2 × 96500 C deposit Cl₂ = 1 mol

 ∴ 18 C will deposit Cl₂ = 1 mol × 18 C × 6.022 × 10²³ / 2 × 96500 C molecules mol⁻¹

FINAL RESULT = 5.616 × 10¹⁹ molecules.

Hope it helps you !!