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 36.00 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.

Expert Solution

Let mass of NaCl be m gram
Then mass of KBr will be (0.3146-m) g

Molar mass of NaCl= 58.5 g/mol
number of moles of NaCl = mass / molar mass
= m/58.5

Molar mass of KBr= 199 g/mol
number of moles of NaCl = mass / molar mass
= (0.3146- m) /199
So, total number of moles of Cl- and Br - = m/58.5 + (0.3146- m) /199

Number of moles of AgNO3 required = M*V
= 0.08765*0.036 (since volume is 36 mL= 0.036 L)
=3.16*10^-3 mol

Number of moles of AgNo3 = number of moles of Cl- and Br
3.16*10^-3 = m/58.5 + (0.3146- m) /199
3.16*10^-3 = (199*m+ 58.5* (0.3146- m)) /(199*58.5)
3.16*10^-3 = (199*m+ 18.4 - 58.5*m) /(11641.5)
36.79 = 140.5 m + 18.4
m = 0.1309 g

mass % of NaCl = mass of NaCl * 100 / total mass
= 0.1309 * 100 / 0.3146
= 41.6 %
Answer: 41.6 %

queen_honey_blossom answered 1 year ago

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