Question

1.52 gm of a trivalent metal (M) was deposited at cathode by passing a current
of 2.5 ampere through its salt solution (metal sulphate) for 30 minutes. What is
the atomic mass of M? (Ans: 97.78 amu)​

Answer 1

$\huge \tt97.78 \: \: \: amu$

Answer 2

Given:

Weight 'w' of the metal = 1.52 g

Current = 2.5 amp.

Time = 2.5 × 30 × 60 sec

Valency = 3

To find :

The atomic mass of metal (M)

Solution :

According to Faraday's first law of electrolysis:

• The mass of a substance deposited at any electrode is directly proportional to the amount of charge passed.

• The constant of proportionality Z is called the electro-chemical equivalent of the substance

Electrochemical equivalent Z

$= \frac{w}{i \times t}$

Here,w = weight of metal

i = current in amp

t = time

$= \frac{1.52}{2.5 \times 30 \times 60}$

$= 3.37 \times 10 {}^{ - 4}$

As we know,

Valency = atomic weight of metal / equivalent weight

3 = atomic weight of metal / 3.37 ×10^-4 × 9650

We Know, Equivalent weight = Z × F

( 1F = 9650)

$atomic \: weight \: m = 3 \times 3.37 \times 10 {}^{ - 4}×9650$

$= 97.78$

Thus, the atomic mass of Metal ( M) deposited at cathode is 97.78 amu.