What quantity of aluminium is deposited when a current of 10A is passed through a solution of an aluminium salt for 1930s? [Al = 27, F = 96500 C mol -1]

What quantity of aluminium is deposited when a current of 10A is passed through a solution of an aluminium salt for 1930s?

[Al = 27, F = 96500 C mol -1]

Answer Details

The quantity of aluminium deposited can be calculated using the formula: Quantity of metal deposited = (I x t x m) / (n x F) where I is the current in amperes, t is the time in seconds, m is the molar mass of the metal, n is the number of electrons required to reduce one mole of the metal ions, and F is the Faraday constant. In this case, I = 10A, t = 1930s, m = 27g/mol, n = 3 (since aluminium ions have a 3+ charge and require three electrons to be reduced to metallic aluminium), and F = 96500 C/mol. Plugging these values into the formula, we get: Quantity of metal deposited = (10 x 1930 x 27) / (3 x 96500) = 1.8 g Therefore, the answer is option B: 1.8 g.