Students performed a procedure similar to part ii of this experiment (analyzing juices for vitamin c content) as described in the procedure section. three 10.00ml samples of juice were titrated with dcp that had a standardized concentration of 9.98x10-4m. the three titrations took an average of 16.34ml of dcp. calculate the mass (in mg) in 50.00ml of juice. (mm ascorbic acid = 176.124 g/mol)

Explanation:

Hello,

In this case, the titration is based of the equality between the moles of the titrant and the analyte, then, in terms of molarity we define it as shown below:

As the ascorbic acid is the substance reacting with the DCP, we infer that the 16.34 mL of DCP completely neutralized the ascorbic acid; in such a way, solving for the molarity of the neutralized acid, one obtains:

Finally, by using the molar MASS of the ascorbic acid and applying the proper units conversion, the result is shown below:

Therefore, 14.36 mg of ascorbic acid are contained into 50.00mL of juice.

Best regards.