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 62.9mL 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 48.8mL of titrant is added.

Part A) If the sample of chips used to make the filtrate weighed 82.5g , how much NaCl is present in one serving (105g ) of chips? Express your answer to three significant figures and include the appropriate units. (in grams)

Explanation / Answer

There is a very simple method to do , which is by molar proportionality:
The 0.500g KCl required 64.0ml of the AgNO3 soltion.
The chip extract took 45.2ml of AgNO3

This is equal to 45.2/64.0*0.500 = 0.353g KCl

But the chip extract was not KCl, but NaCl
Molar mass KCl = 74.55 g/mol
Molar mass NaCl = 58.44 g/mol

0.353g KCl = 0.353*58.44/74.55 = 0.276g NaCl in sample.
Sample mass = 95g chips.
NaCl in 155g chips = 155/95*0.276 = 0.451 g NaCl per serving.

There is another long and complicated way of doing this by using mol of KCl, mol NaCl mol AgNO3, molarity of the AgNO3 soution etc, but I think that this is an easy way of ding it. Just imagine how much simpler it would have been if the chemist had used NaCl instead of KCl in the first titration

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