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

0.122 mg or 0.0122%

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%

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.