As a food chemist for a major potato chip company, you are responsible for deter

Question

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 64.1 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 49.2 mL of titrant is added.

If the sample of chips used to make the filtrate weighed 75.5 g , how much NaCl is present in one serving (125 g ) of chips?

Explanation / Answer

Balanced reaction:
KCl(aq) + AgNO3(aq) AgCl(s) + KNO3(aq)

0.500 g of KCl = (0.500 g)/(74.5513 g/mol) = 6.707 x 10^-3 moles KCl

Molarity of AgNO3 solution = (6.707 x 10^-3 moles KCl)*(1 mol AgNO3/1 mol KCl)/((64.9 mL)*(1 L/1000 mL))
Molarity of AgNO3 solution = 0.1033 M AgNO3

"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 49.2 mL of titrant is added."

Moles of NaCl in sample = (0.1033 M AgNO3)*((49.2 mL)*(1 L/1000 mL)*(1 mol NaCL/1 mol AgNO3)
Moles of NaCl in sample = 5.08236 x 10^-3 moles NaCl

mass of NaCl in sample = (moles NaCl)*(molar mass NaCl) = (5.08236 x 10^-3 moles NaCl)*(58.44 g/mol)
mass of NaCl in sample = 0.2970 g

"If the sample of chips used to make the filtrate weighed 75.5 g , how much m NaCl is present in one serving (125 g) of chips?"

% (w/w) NaCl in sample = 100%*(0.2970 g)/(75.5 g) = 0.3934 % NaCl

Mass of NaCl in 125 g = (125 g)*(0.3934 % NaCl) = 0.4917 g of NaCl

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