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, NaCI 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 places You then analyze the chip filtrate for C concentration using the Mohr method.

First you prepare a solution of silver nitrate, AgNO3, and titrate it against 0 500 g of KCI using the Mohr method You find that t takes 62 5 mL of AgNO, ttrant 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 47.2 mL of titrant is added. 

Part A 

if the sample of chips used to make the filtrate weighed 94.0 g how much NaCl is present in one serving (115g) of chips

Answer 1

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

no of moles of KCl   = W/G.M.Wt
                     = 0.5/74.55   = 0.006707 moles of KCl
no of moles of KCl = no of moles of AgNO3
molarity of AgNO3 = no of moles/volume in L
                   = 0.006707/0.0625   = 0.107312M
47.2 ml of AgNO3 solution to precipiate all of the chloride.
no of moles of Cl^-   = molarity * volume in L
                       = 0.107312*0.0472 = 0.005065 moles
no of moles of Cl^- = no of moles of NaCl
mass of NaCl   = no of moles * gram molar mass
               = 0.005065*58.5   = 0.2963g of NaCl
    115/94 = 1.2234 chips
    1.2234*0.2963 = 0.3625g NaCl in 115g chips