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 64.6 mL of AgNO3titrant 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 47.9 mLof titrant is added.

If the sample of chips used to make the filtrate weighed 83.0 g , how much NaCl is present in one serving (115 g ) of chips?

Express your answer to three significant figures and include the appropriate units.

Answer 1

Solution :-

Balanced reaction equations are as follows.

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

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

Lets first calculate the moles of the KCl

Moles = mass / molar mass

Moles of KCl = 0.500 g / 74.55 g per mol

                        = 0.006707 mol KCl

1 mol KCl gives 1 mol Cl^-

Therefore using the moles of KCl we can find the moles of AgNO3

0.006707 mol KCl * 1 mol AgNO3 / 1 mol KCl = 0.006707 mol AgNO3

Now lets calculate the molarity of the AgNO3 using the moles and volume of AgNO3 used to titrate the KCl

Molarity = moles / volume in liter

Volume of AgNO3 = 64.6 ml = 0.0646 L

Molarity of AgNO3 = 0.006707 mol / 0.0646 L

                                   = 0.1038 AgNO3

Now using the molarity and volume of AgNO3 used for the titration for the chips filtrate we can find the moles of AgNO3 and then using these moles we can find moles of Cl^-

moles = molarity x volume in liter

volume of AgNO3 = 47.9 ml = 0.0479 L

moles of AgNO3 = 0.1038 mol per L * 0.0479 L

                              = 0.004972 mol AgNO3

Now using this moles of AgNO3 lets calculate the moles of NaCl

0.004972 mol AgNO3*(1 mol Cl^- / 1mol AgNO3)*(1 mol NaCl / 1 mol Cl^-) = 0.004972 mol NaCl

Converting moles of NaCl to its mass

Mass= moles x molar mass

Mass of NaCl = 0.004972 mol * 58.0443 g per mol

                          = 0.290 g NaCl

So, 0.290 g NaCl present in the filtrate that is in 83.0 g chips

Now lets calculate it for the one serving (115 g chips)

(0.290 g NaCl * 115 g chips) / (83.0 g chips) = 0.402 g NaCl

Therfore one serving of the chips contains 0.402 g NaCl salt.