Question

A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved in water. The resulting solution required 47.50 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.

Answer #1

mass of KBr = m₂

m₁ + m₂ = 0.3146

moles:

NaCl: n = mass / molar mass = m₁/ 58.443 mol

KBr: n = mass / molar mass = m₂/ 119.002 mol

AgNO₃: n = molarity *volume = (0.08765 mol/L)(47.50 mL x 1L/1000mL)
= 0.004163 mol

net ionic equation:

2Ag⁺(aq) + Cl⁻(aq) + Br⁻(aq) → AgCl(s) + AgBr(s)

m₁/ 58.443 + m₂/ 119.002 = 0.004163

solving system of equations:

m₁ + m₂ = 0.3146

m₁/ 58.443 + m₂/ 119.002 = 0.004163

m₁ = 0.1744g, m₂ = 0.1401g

``````````````````````````````````````...

mass % NaCl in the mixture:

mass% = [m₁ /( m₁ + m₂ ) ]100%

mass% = [0.1744 /( 0.3146 ) ]100%

mass% = 55.43%

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