Question

You enter the lab to analyze the chip filtrate for Cl?. 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 62.7mL of AgNO3 titrant to fully 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 47.9mL of titrant is added.

If the sample of chips used to make the filtrate weighed 80.0g , how much NaCl is present in one serving (155g ) of chips?

Answer 1

Molarity of your AgNO3 if it takes 63.9 mL of AgNO3 titrant against 0.500 grams of KCl

moles of KCl = 0.500/74.551 g/mol = 6.707x10^-3 moles KCl

since AgNO3 & KCl react 1 mole to 1 mole:

6.707x10^-3 moles KCl reacts with 6.707x10^-3 moles AgNO3

Molarity of 62.7 ml solution of AgNO3 = 6.707x10^-3/0.0627 = 0.107 M AgNO3

80 g of chips

moles in 47.9 mL of AgNO3 = 0.0479*0.107 = 5.1237x10^-3 moles of AgNO3

NaCl reacts with AgNO3 in a 1 mole to 1 mole ratio:

5.1237x10^-3 mol AgNO3 reacted with 5.1237x10^-3 moles of NaCl

5.1237x10^-3*58.44 g/mol = 0.3 g NaCl

salt in a 155 gram serving:

155*0.3 grams of NaCl / 80 grams analyzed= 0.58 grams of salt