Solution:

Given:
Number of electrons passed through the solution, n = 6 × 10^22
Electrolyte used, CuSO4
Electrodes used, platinum electrodes

To find: Total loss in weight of the solution

Formula:
The loss of weight is given by the formula:
Loss of weight = (m2 - m1)
where m1 is the initial mass of the solution and m2 is the final mass of the solution.

Calculation:
To calculate the loss in weight of the solution, we need to find the amount of copper deposited on the cathode and the amount of copper sulfate decomposed at the anode.

At the cathode:
Cu2+ + 2e- → Cu (reduction)
The number of electrons required to reduce one mole of Cu2+ ions to Cu is 2.
The atomic mass of Cu is 63.5 g/mol.

The amount of copper deposited on the cathode can be calculated as follows:
Number of moles of Cu deposited = (n/2NA)
= (6 × 10^22)/(2 × 6.022 × 10^23)
= 0.05 moles

Mass of Cu deposited = number of moles of Cu deposited × atomic mass of Cu
= 0.05 × 63.5
= 3.175 g

At the anode:
H2O → 2H+ + 1/2O2 + 2e- (oxidation)

The number of electrons required to oxidize one mole of water to oxygen gas is 4.
The atomic mass of oxygen is 16 g/mol.

The amount of oxygen gas formed at the anode can be calculated as follows:
Number of moles of O2 formed = (n/4NA)
= (6 × 10^22)/(4 × 6.022 × 10^23)
= 0.025 moles

Mass of O2 formed = number of moles of O2 formed × atomic mass of O2
= 0.025 × 16
= 0.4 g

The amount of sulfuric acid formed at the anode can be calculated as follows:
Number of moles of H+ ions formed = (n/2NA)
= (6 × 10^22)/(2 × 6.022 × 10^23)
= 0.05 moles

Number of moles of H2SO4 formed = 0.05/2
= 0.025 moles

Mass of H2SO4 formed = number of moles of H2SO4 formed × molar mass of H2SO4
= 0.025 × (2 × 1 + 32 + 4 × 16)
= 0.25 g

Total loss in weight of the solution = (m2 - m1)
= (3.175 + 0.4 + 0.25) - m1
= 3.825 - m1

Answer:
The total loss in weight of the solution is 3.825 g, which is approximately equal to 4 g.
Therefore, the correct option is (a) 4 grams.