Chemistry 2010 JAMB Past Questions - Question 25: What quantity of aluminium is deposited when a current of 10A is passed through a solution of an aluminium salt for 1930s? .

Correct Answer

B

Explanation

1. **Molarity of the aluminum salt solution:** This information defines the concentration of aluminum ions available for deposition.
2. **Specific aluminum salt used:** Different salts like AlCl? or Al?(SO?)? have different molar masses, affecting the amount of aluminum deposited per mole of the salt.
3. **Electrolysis time:** Although you mentioned "1930s," it's unclear if this refers to time in seconds (s), minutes (min), or hours (h). Knowing the exact time duration is crucial for the calculation.

Once you provide these missing details, I can use Faraday's law of electrolysis to calculate the mass of aluminum deposited:

**m = (ItzF) / M**

where:

* **m** is the mass of deposited metal (aluminium in this case)
* **I** is the current (10 A)
* **t** is the electrolysis time (unknown)
* **z** is the number of electrons transferred per metal ion (for Al³?, z = 3)
* **F** is the Faraday constant (96,485 C/mol)
* **M** is the molar mass of aluminum (27 g/mol)