A constant current was passed through a solution of $$AuCl_4^-$$ ion between gold electrodes. After a period of $$10.0$$ minutes, the increase in mass of cathode was $$1.314$$ g. The total charge passed through the solution is ______ $$\times 10^{-2}$$ F. (Given atomic mass of Au = 197)

We need to find the total charge passed through a solution of $$AuCl_4^-$$ during electrolysis. The cathode gained 1.314 g in mass, the electrolysis lasted 10.0 minutes, and the atomic mass of Au is 197 g/mol.

In $$AuCl_4^-$$, gold is in the $$+3$$ oxidation state since each Cl is $$-1$$, making Au $$+3$$. At the cathode, gold is deposited by reduction according to $$Au^{3+} + 3e^- \rightarrow Au$$, so 3 Faradays of charge deposit 1 mole (197 g) of gold.

The moles of Au deposited are given by $$\text{Moles of Au} = \frac{1.314}{197} = 0.00667 \text{ mol}$$.

Since 3 moles of electrons are required per mole of Au, the charge in Faradays is $$\text{Charge} = 3 \times 0.00667 = 0.02 \text{ F}$$, which can be expressed as $$0.02 \text{ F} = 2 \times 10^{-2} \text{ F}$$.

Therefore, the answer is 2.