Solution:

Given, charge used = 0.1 F

We know that, the amount of substance liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

The amount of substance liberated is given by the formula,

W = ZIt

where W is the amount of substance liberated in grams, Z is the electrochemical equivalent, I is the current in amperes and t is the time in seconds.

In this case, since the charge used is given, we can find the time required to pass this charge using the formula,

Q = It

where Q is the charge in coulombs.

Therefore, t = Q/I = 0.1/1 = 0.1 seconds

Now, we can find the amount of substance liberated using the formula,

W = ZIt

For K2SO4, Z = M/2F = 39/2 = 19.5 gm/coulomb

Therefore, W = 19.5 x 0.1 x 2 = 3.9 gms

However, during electrolysis, water will also be electrolyzed to produce hydrogen and oxygen gas. Since we are using platinum electrodes, these gases will not react with the electrodes and will escape into the atmosphere. Therefore, the total loss in mass of the solution will be the sum of the mass of the substance liberated and the mass of the gases produced.

The mass of hydrogen and oxygen produced can be calculated using Faraday's laws of electrolysis.

For hydrogen gas,

2H+ + 2e- -> H2

1 mole of electrons produces 1 mole of hydrogen gas

1 mole of electrons = 96500 coulombs

1 mole of hydrogen gas = 2 gm

Therefore, the mass of hydrogen produced = (0.1 x 2)/96500 = 2.07 x 10^-4 gms

Similarly, for oxygen gas,

2H2O -> 2H+ + O2 + 2e-

1 mole of electrons produces 1/2 mole of oxygen gas

1 mole of electrons = 96500 coulombs

1 mole of oxygen gas = 32 gm

Therefore, the mass of oxygen produced = (0.1 x 32)/2 x 96500 = 1.66 x 10^-3 gms

Therefore, the total loss in mass of the solution is given by,

3.9 + 2.07 x 10^-4 + 1.66 x 10^-3 = 3.91 gms

Thus, the correct option is (4) 4.7gms.