Question

A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved in water. The resulting solution required 48.80 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.​

______% NaCl

Answer 1

Mass of NaCl = m₁
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)(48.80 mL x 1L/1000mL) = 0.004277 mol

Net ionic equation:
2Ag⁺(aq) + Cl⁻(aq) + Br⁻(aq) → AgCl(s) + AgBr(s)
m₁/ 58.443 + m₂/ 119.002 = 0.004277

Solving system of equations:
m₁ + m₂ = 0.3146 -----(1)
m₁/ 58.443 + m₂/ 119.002 = 0.004277 -------(2)

Now, m₁/ 58.443 + m₂/ 119.002 = 0.004277

=> 119.002 m₁ + 58.443 m₂ /58.443 x 119.002 =0.004277

=>119.002(0.3146 -m₂) +58.443 m₂ =29.7458 ,using equation (1) m₁=0.3146 -m₂

=> 60.559 m₂ =7.6922

=> m₂ =0.1270 g

Now, m₁ + m₂ = 0.3146 =>m₁ =0.3146-0.1270 =0.1876 g

Mass % NaCl in the mixture:
Mass% = [m₁ /( m₁ + m₂ ) ]100%
Mass% = [0.1876 /( 0.3146 ) ]100%
Mass% = 59.63%