First you prepare a solution of silver nitrate, AgNO3, and titrate it against 0.

Question

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 61.3 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 47.7 mL of titrant is added. If the sample of chips used to make the filtrate weighed 82.5 g , how much NaCl is present in one serving (135 g ) of chips?

Explanation / Answer

First, determine the molarity of the AgNO3 solution.

Molarity = number of moles / volumes in L

The number of moles are equals to the numbe of moles of KCl.

Number of moles of KCl= amount in g/ molar mass

= 0.500 g / 74.55 g/ mol

=6/.71*10^-3 moles KCl

Molarity of AgNO3 solution.

Molarity = number of moles / volumes in L

Volume in L = 61.3 ml* 1 .0 L/1000 ml= 0.0613 L

Molarity of AgNO3= 6.71mole*10^3/0.0613L

= 0.1094 M

47.7 ml of the AgNO3 solution has the same number of moles as the salt in the the filtrate weighed 82.5 g sample:
0.1094 mole/L x 0.0477L

= 5.22*10^-3 moles NaCl in the sample.

Now calculate the mass of 5.22*10^-3 moles NaCl:

5.22*10^-3 moles NaCl x molar mass

5.22*10^-3 moles NaCl *58.44 g

= 0.305 g NaCl in 82.5 g sample

Now calculate the amount of NaCl For a 135g serving:
= 0.305g x (135/82.5)

= 0.499 g or 0.5 NaCl in a 135g serving

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