A protein solution (0.2 ml) of unknown concentration was diluted with 0.8 ml of water. To 0.5 ml of this diluted solution 4.5 ml of biuret reagent was added and the color allowed to develop. The absorbance of this mixture taken in a test tube of 1cm diameter at 540 nm was observed to be 0.20. 0.5 ml of BSA (4 mg/ml) solution plus 4.5 ml of biuret gave an absorbance of 0.20 when measured as above. What is the protein concentration (mg/ml) in the undiluted solution?​

Correct option is A

Explanation-

Unknown sample:
      0.2 ml protein solution (unknown concentration) + 0.8 ml water = 1.0 ml
      0.5 ml of this diluted sample + 4.5 ml biuret → Absorbance = 0.20
Standard sample:
      0.5 ml of BSA (4 mg/ml) + 4.5 ml biuret → Absorbance = 0.20

Step 1: Protein in the Standard Sample
             Volume taken = 0.5 ml
             Concentration = 4 mg/ml
So, protein content =
                           $$0.5 \, \text{ml} \times 4 \, \text{mg/ml} = 2 \, \text{mg}$$

Step 2: Protein in the Unknown Sample
From the diluted solution (total = 1.0 ml), 0.5 ml is taken. The 1.0 ml diluted solution came from 0.2 ml of the original unknown solution (plus 0.8 ml water).
Let original concentration be x mg/ml
           Protein in 0.2 ml = $$0.2 \, \text{ml} \times x \, \text{mg/ml} = 0.2x \, \text{mg}$$

Since 0.5 ml is taken from this 1 ml diluted solution:

$$\text{Protein in test tube} = \frac{0.5}{1.0} \times 0.2x = 0.1x \, \text{mg}$$

Step 3: Equating to the Standard
Since both absorbances are 0.20, amounts of protein are equal:​​

                                       $$0.1x = 2 \Rightarrow x = \frac{2}{0.1} = \boxed{20 \, \text{mg/ml}}$$​​

So, the correct answer is option a :  20 mg/ml