The mass of copper deposited when passing an electric current through a copper sulphate solution can be calculated using Faraday's laws of electrolysis. The first law states that the amount of a substance deposited or liberated at an electrode during electrolysis is proportional to the amount of electricity (charge) passed through the electrolyte.

To find the mass of copper deposited, we use the formula:

$$ m = \frac{M \times I \times t}{n \times F} $$

Where:

• $ m $ is the mass of the substance deposited (in grams),
• $ M $ is the molar mass of the substance (in g/mol),
• $ I $ is the current (in amperes, A),
• $ t $ is the time electricity is passed through the solution (in seconds),
• $ n $ is the number of moles of electrons required to deposit or dissolve 1 mole of the substance (valence number, which is 2 for copper in copper sulphate solution as copper ions are $\mathrm{Cu^{2+}}$),
• $ F $ is the Faraday constant, approximately $ 96487 \mathrm{C/mol} $ of electrons.

Given:

• $$ M = 63 \mathrm{~g/mol} $$
• $$ I = 9.6487 \mathrm{~A} $$
• $ t = 100 \mathrm{~s} $
• $ n = 2 $
• $$ F = 96487 \mathrm{~C/mol} $$

Substitute these values into the formula:

$$ m = \frac{63 \mathrm{~g/mol} \times 9.6487 \mathrm{~A} \times 100 \mathrm{~s}}{2 \times 96487 \mathrm{~C/mol}} $$

Calculate the value of $ m $:

$$ m = \frac{63 \times 9.6487 \times 100}{2 \times 96487} $$

$$ m = \frac{60828.21}{192974} $$

$$ m \approx 0.315 \mathrm{~g} $$

Hence, the mass of copper deposited is approximately 0.315 g, making Option B the correct answer.