Question

3.8g of metal Mis deposited at the cathode by passing 0.2F of electricity through the solution of met
chloride. The formula of the metal chloride (atomic mass of M=57)
A) MCI
B) MCI,
C) MCI,
D) MCI,​

Answer 1

Given:

3.8g of metal Mis deposited at the cathode by passing 0.2F of electricity through the solution of metal chloride.

To find:

Formula of metal chloride.

Calculation:

Let the following equation tale place:

${M}^{n + } + ne \: \rightarrow \: M$

So, n F of electricity will deposit 1 mole of M

=> n F of electricity will deposit 57 g of M

=> 0.2 F of electricity will deposit

$= \dfrac{57}{n} \times 0.2 \: g \: of \: M$

Now , this value should be equal to 3.8 grams.

$\therefore \dfrac{57}{n} \times 0.2 = 3.8$

$= > \dfrac{57}{n} \times 2 = 38$

$= > \dfrac{3}{n} \times 2 = 2$

$= > n = 3$

So, valency of M = 3.

So, formula of metal chloride will be:

$\boxed{ \bf{required \: formula = MCl_{3}}}$