Question

1. One snack food packaging contains 37 grams of dried vegetable. The student takes the entire contents of the package and grinds it using pestle and mortar. After the sample is prepared, the student takes exactly 1.00 grams of sample and dissolves it in adequate volume of water. She then adds 5 drops of indicator and titrates to the end-point using 0.100 M AgNOs solution. The volume of AgNOa needed for titration is 10.50 mL. a. What is the amount of salt in the package? (recommended daily allowance) of sodium is 904 mg. What is the percent RDA of sodium b. The RDA intake if a person eats the content of the package? Na 23.

Answer 1

Ans 1 :

a)

The reaction is given as :

NaCl + AgNO3 = AgCl + NaNO3

Number of moles of AgNO3 = molarity x volume (L)

= 0.100 x 0.01050

= 0.00105 moles

So number of moles of salt will also be 0.00105 moles in 1 gram

The number of moles of salt in package = 37 x 0.00105 = 0.03885 moles

Mass in grams = 0.03885 x molar mass of NaCl

= 0.03885 x 58.44

= 2.27 grams

b)

Number of moles of sodium present in package is 0.03885

Mass of sodium will be : 0.03885 x 23

= 0.89355 g or 893.55 mg

So percent of RDA intake = (893.55 / 904) x 100

= 98.8 %