Question

7. A 5.00 g sample of spinach was extracted with 1% oxalic acid and the extract was diluted to 50.0 ml. The titration of 1.00 ml DCPIP required 20.6 ml and 19.8 ml in duplicate titrations. The titration required 7.8 ml and 8.6 ml of a 0.030 mg/ml standard solution of L-ascorbic acid. How much vitamin C was in 5 g of spinach

Answer 1

0.122 mg or 0.0122%

Step-by-step explanation:

We first calculate average of ascorbic acid

= 7.8 + 8.6 /2

= 17.4/2

= 8.2

8.2 ml contains 0.03 x 8.2

= 0.246 mg of ascorbic acid.

1 ml of dcpip reacts with 0.246 ascorbic acid

1 mol of dcpip required 20.6 ml and 19.8

Average = 19.8 + 20.6 / 2

= 40.4/2

= 20.2ml

If 20.2 ml = 0.246 ascorbic acid

50 ml = 0.246 x 50 / 20.2

= 0.6089 mg is in 50 ml extract

5 mg of spinach is what was used to make extract

= 0.6089/5

= 0.122 mg

Vitamin c is made up of ascorbic acid as it's composition. So Vitamin c in spinach = 0.122 mg or 0.0122%