To solve this problem, we need to use the concept of stoichiometry and the law of conservation of mass. Let's break down the problem into steps:

Step 1: Calculate the moles of AgCl and AgBr
We are given the weight of the precipitate of AgCl and AgBr, which is 0.4066 g. We can assume that the weights of AgCl and AgBr are equal since they are both precipitated together. Therefore, the weight of AgCl is 0.4066 g / 2 = 0.2033 g.

To calculate the moles of AgCl and AgBr, we need to divide their weights by their respective molar masses. The molar mass of AgCl is 143.32 g/mol and the molar mass of AgBr is 187.77 g/mol.

Number of moles of AgCl = 0.2033 g / 143.32 g/mol ≈ 0.00142 mol
Number of moles of AgBr = 0.2033 g / 187.77 g/mol ≈ 0.00108 mol

Step 2: Calculate the moles of AgBr converted to AgCl
When AgBr is heated in a current of chlorine, it is converted to AgCl. We are given that the mixture loses 0.0725 g in weight. This weight loss is due to the loss of AgBr, which is converted to AgCl.

To calculate the moles of AgBr converted to AgCl, we need to divide the weight loss by the molar mass of AgBr.

Number of moles of AgBr converted to AgCl = 0.0725 g / 187.77 g/mol ≈ 0.00039 mol

Step 3: Calculate the moles of chlorine
From the stoichiometry of the reaction, we know that 1 mole of AgBr is converted to 1 mole of AgCl when heated in a current of chlorine. Therefore, the moles of chlorine consumed is equal to the moles of AgBr converted to AgCl.

Number of moles of chlorine = 0.00039 mol

Step 4: Calculate the % of chlorine in the original solution
To calculate the % of chlorine in the original solution, we need to divide the moles of chlorine by the moles of the original solution and multiply by 100.

% of chlorine = (0.00039 mol / (0.00108 mol + 0.00142 mol)) * 100 ≈ 6.15%

Therefore, the % of chlorine in the original solution is approximately 6.15%.