Please help me. Thanks!!!! As a food chemist for a major potato chip company, yo

Question

Please help me. Thanks!!!!

As a food chemist for a major potato chip company, you are responsible for determining the salt content of new potato chip products for the packaging label. The potato chips are seasoned with table salt, NaCl. You weigh out a handful of the chips, boil them in water to extract the salt, and then filter the boiled chips to remove the soggy chip pieces. You then analyze the chip filtrate for Cl concentration using the Mohr method.

First you prepare a solution of silver nitrate, AgNO3, and titrate it against 0.500 g of KCl using the Mohr method. You find that it takes 63.3 mL of AgNO3 titrant to reach the equivalence point of the reaction.

You then use the same silver nitrate solution to analyze the chip filtrate in a Mohr reaction, finding that the solution yields a rusty brown precipitate when 45.1 mL of titrant is added.

Part A

If the sample of chips used to make the filtrate weighed 79.0 g , how much NaCl is present in one serving (115 g ) of chips?

Explanation / Answer

Summary of data:

We have to determine salt content.

Salt is NaCl.

Method used to Mohr method.

Solution prepared of AgNO3

Titrated against 0.500 g KCl   (using Mohr Method.)

Volume of AgNO3 used in titration = 63.3 mL

Again AgNO3 is titrated and titrant used = 45.1 mL

Weight of sample = 79.0 g

One serving of chips weigh 115 g

We have to find mass of NaCl in each serving.

Solution:

Strategy:

First we find Molarity of AgNO3 in first titration with KCl.

The we use this molarity to get moles in second titration .

From the moles of AgNO3, by writing reaction with NaCl we find amount of moles of NaCl.

That mol is present in 79.0 g . Then we find moles of NaCl in 115 g

The obtained moles are used to get mass of NaCl.

Solution:

Reaction : Titration of AgNO3 Vs KCl

AgNO3 (aq) + KCl (aq) --- > AgCl (s) + KNO3 (aq)

Mol ratio : AgNO3 : KCl is 1:1

Moles of KCl = 0.500 g x 1 mol / 74.551 g

= 0.006707 mol KCl

Moles of AgNO3 = 0.006707 mol KCl x 1 mol AgNO3/ 1 mol AgNO3

= 0.006707 mol AgNO3

[ AgNO3 ]= Mole / volume in L

= 0.006707 mol / 63.3 x 10E-3 L

= 0.1059 M

Titration of AgNO3 with NaCl

AgNO3 (aq) + NaCl (aq) ---- > AgCl(s) + NaNO3 (aq)

Mol ratio : AgNO3 : NaCl is 1:1

Moles AgNO3 used = Molarity x volume in L

= 0.1059 M x 0.0451 L

= 0.004779 mol AgNO3

n NaCl = 0.004779 mol AgNO3 x 1 mol NaCl / 1 mol AgNO3

= 0.004779 mol NaCl

n NaCl present in 115 g = 115 g serve x 0.004779 mol/79.0 g

= 0.006959 mol NaCl

Mass of NaCl = 0.006959 mol NaCl x molar mass of NaCl

= 0.006959 mol NaCl x 58.44 g /mol

= 0.41 g

Mass of NaCl in each serve = 0.41 g

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