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 64.9 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 48.2 mL of titrant is added.

Part A

If the sample of chips used to make the filtrate weighed 83.0 g , how much NaCl is present in one serving (155 g ) of chips?

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

Answer 1

The reaction between KCl & AgNO3 is given by      AgNO3 + KCl AgCl + KNO3

Molar mass of KCl = 39+35.5 = 74.5 g/mol

MOlar mass of AgNO3 = 108 + 14+(3x16) = 170 g/mol

According to the balanced Equation ,

1 mole of KCl reacts with 1 mole of AgNO3

                          or

1x74.5 g of KCl reacts with 1x170 g of AgNO3

0.50 g of KCl reacts with M g of AgNO3

M = ( 1x170x0.50 ) / ( 1x74.5)

    = 1.14 g of AgNO3

Number of moles of AgNO₃ , n = mass/molar mass

                                                  = 1.14 g / 170 (g/mol)

                                                  = 6.7x10 ^-3 mol

So molarity of AgNO3 , M = number of moles / volume in L

                                         = 6.7x10 ^-3 mol / (64.9 x 10 ^-3 L )

                                         = 0.103 M

In the chips there is NaCl & this NaCl solution is titrated with AgNO3 solution so the reaction is

     NaCl + AgNO3 AgCl + NaNO3

Molar mass of AgCl = 108+35.5 = 143.5 g/mol

Molar mass of NaCl = 23+35.5=58.5 g/mol

From the balanced equation ,

1 mole of NaCl produces 1 mole of AgCl

                              OR

1x58.5 g of NaCl produces 1x143.5 g of AgCl

N g of NaCl produces 83.0 g of AgCl

N = ( 1x58.5x83.0) / ( 1x143.5)

   = 33.8 g of NaCl

Therefore the mass of NaCL present in one serving of sample is 33.8 g