1. One snack food packaging contains 37 grams of dried vegetable. The student ta

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 AgNO3 solution. The volume of AgNO3 needed for titration is 10.50 mL. a. What is the amount of salt in the package? b. The RDA of sodium is 904 mg. What is the percent RDA of sodium intake if a person eats the contert of the package? Na 23

Explanation / Answer

I don't have the idea what types of salts are present in dried vegetables. From the 2nd part of the question, I guessed it be Sodium Chloride(NaCl). If it is the case, then the following reaction is taking place when you add your AgNO3.

NaCl + AgNO3 = AgCl + NaNO3

Provided, AgNO3 solution= 0.100 M

Volume of required AgNO3 for titration= 10.50 ml

Hence, from balanced equation,

Amount of Prepared AgCl= 0.100mol x 10.50ml = 1.05 mmol (mili-mole)

In other word, amount of NaCl reacted= 1.05 mmol

(note the stoichiomerty of the reactants and products are same i.e. 1)

Now 1.05 mmol of NaCl= 1.05 x MW of NaCl = 1.05 (mili-mol)x 58.4 (gm/ mol)= 61.32 mg of NaCl

So, 61.32 mg NaCl is within 1g of sample.

So, the amount of salt in package= 61.32x 37= 2.269 g

* Note: if there is divalent salt like CaCl2, then simply write the equation and follow the same steps. But don't forget to multiply the mmol by the stoicheometric coefficient.

** The 2nd part i.e. RDA, is the concept of nutrition. I think, it'd better to solve it by any specialist in that field.

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