Question

A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved in water. The resulting...

A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved in water. The resulting solution required 47.20 mL of 0.08765 M AgNO3(aq) to precipitate the Cl–(aq) and Br–(aq) as AgCl(s) and AgBr(s). Calculate the mass percentage of NaCl(s) in the mixture.

Homework Answers

Answer #1

1 mol NaCl = 1 mol AgNo3

1 mol NaBr = 1 mol AgNo3

total no of mol of AgNo3 reacted = 47.2*0.08765/1000 =

                                  = 0.00414 mol

mass of NaCl = x*58.5

mass of NaBr = (0.00414-x)*103

0.3146 = x*58.5 + (0.00414-x)*103

x = 0.0025 mol

mass percentage of NaCl(s) in the mixture = 0.0025*58.5/0.3146*100

        = 46.5%

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