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)

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Mariam Spence

Chemistry

14 July, 01:24

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)

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Carley Goodman · Answer 1

Students performed a procedure similar to Part II of this experiment (Analyzing Juices for Vitamin C Content) as described in the procedure section. Given that molarity of DCP is 9.98x10-4 M, it took 16.34 ml of DCP to titrate 10 mL of sample.

Amount of ascorbic acid = 0.050 L sample (0.01634 L DCP/0.01 L sample) (9.98x10-4 mol DCP/L DCP) (1 mol Ascorbic acid / 1mol DCP) (176.124 g/mol) (1000mg/1g) = 14.36 mg ascorbic acid