Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The density of $\text{Cu}$ is $\text{8}\text{.94g c}{{\text{m}}^{-3}}\text{ }$. The quantity of electricity needed to plate an area $\text{10cm }\times \text{ 10cm}$ to a thickness of $\text{1}{{\text{0}}^{-2}}\text{cm}$ using $\text{CuS}{{\text{O}}_{4}}$ solution would be:A. 13586 CB. 27172 CC. 40758 CD. 20348 C

Last updated date: 21st Sep 2024
Total views: 431.4k
Views today: 9.31k
Verified
431.4k+ views
Hint: For this problem, we have to use the formula of density first that is the ratio of mass and volume. After calculating the mass, we will use the formula of Faraday's law that is m = EQ where m is the mass of a substance, E is the equivalent weight and Q is the charge.

- In the given question, we have to calculate the quantity of electricity that will be required to plate an area by using a copper sulphate solution.
- It is given that the copper has the density of $\text{8}\text{.94g c}{{\text{m}}^{-3}}\text{ }$ so firstly we will apply density formula to calculate the mass i.e.
$\text{d = }\dfrac{\text{mass}}{\text{volume}}$
$\text{mass = density }\times \text{ volume}$
- So, the mass deposited on the electrode will be:
$\text{mass = 8}\text{.94 }\times \text{ 1 = 8}\text{.94gm}$

- Now, we will calculate the amount of electricity by using the equation of Faraday's law.
- According to the first law of Faraday, mass is directly proportional to the charge i.e.
$\text{m }\propto \text{ Q}$ and the proportional constant will be Z i.e.
$\text{m = ZQ}$ … (1)

- Now, as we know that the Z is also equal to the ratio of equivalent weight and Faraday constant according to the second law of Faraday so equation (1) will become
$\text{m = }\dfrac{\text{E}}{\text{F}}\text{Q}$
- Here, E is the equivalent weight and F is the Faraday's constant which is equal to the 96500.
- Now, by putting all the value we will get:
$8.94\text{ = }\dfrac{63.5\text{ }\times \text{ Q}}{2\text{ }\times \text{ 96500}}$
Q = 27172 C
So, the correct answer is “Option B”.

Note: In the above problem, the equivalent weight is the ratio of the molar mass to the valence factor and the molar mass of copper is 63.5g and the valency of copper will be 2 as it has charge +2 on it that's why the value of the equivalent weight of copper is taken as $\dfrac{\text{63}\text{.5}}{2}$.