What current in ampere will deposit 0.27g of aluminium in 2 hours? [Al = 27,

Explanation

The given question is related to the electrolysis of aluminum. To calculate the current required to deposit 0.27g of aluminum in 2 hours, we need to use Faraday's laws of electrolysis.

First, we need to find the number of moles of aluminum deposited:

Moles of aluminum = mass / molar mass = 0.27g / 27g/mol = 0.01 moles

Now we need to find the charge required to deposit 0.01 moles of aluminum:

1 mole of aluminum requires 3 moles of electrons, as aluminum has a valency of +3. So, 0.01 moles of aluminum will require 0.03 moles of electrons.

The charge required to deposit 0.01 moles of aluminum can be calculated using the Faraday constant (F) which relates the charge (Q) to the amount of substance (n) in moles and the number of electrons per mole:

Q = n * F

Q = 0.03 moles * 96500C/mol = 2895C

Now we need to find the current (I) required to deposit 0.27g of aluminum in 2 hours:

Current (I) = Charge (Q) / Time (t)

Time (t) = 2 hours = 7200 seconds (since 2 hours * 60 minutes/hour * 60 seconds/minute = 7200 seconds)

I = 2895C / 7200s = 0.4021A ? 0.4A

Therefore, the correct answer is Option C: 0.4A. The other options are incorrect because they do not represent the correct current required to deposit 0.27g of aluminum in 2 hours.