Question

A 0.3146 g sample of a mixture of NaCl ( s ) and KBr ( s ) was dissolved in water. The resulting solution required 45.70 mL of 0.08765 M AgNO 3 ( 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

The answer to the question is

The mass percentage of NaCl(s) in the mixture is 49.7%

Step-by-step explanation:

The given variables are

mass of sample of mixture = 0.3146 g

Volume of AgNO₃ required to react comletely with the chloride ions = 45.70 mL

Concentration of the AgNO₃ added = 0.08765 M

The equations for the reactions oare

NaCl(aq) + AgNO₃ (aq) = AgCl(s) + NaNO₃(aq)

AgNO₃ (aq) + KBr (aq) → AgBr (s) + KNO₃

The equation for the reaction shows one mole of NaCl reacts with one mole of AgNO₃ to form one mole of AgCl

Thus 45.70 mL of 0.08765 M solution of AgNO₃ contains
`(45.7)/(1000) (0.08765) = 0.004 moles`

Therefore the sum of the number of moles of Br⁻ and Cl⁻

precipitated out of the solution = 0.004 moles

Thus if the mass of NaCl in the sample = z then the mass of KBr = y

However the mass of the sample is given as 0.3146 g which means the molarity of the solution is 0.004 moles

given by

`(z)/(58.44) + (y)/(119) = 0.004 moles` and z + y = 0.3146

Therefore z = 0.3146 - y which gives

`((0.3146-y))/(58.44) + (y)/(119) = 0.004 moles`

-8.7×10⁻³y +0.54×10⁻³ = 0.004

or 8.7×10⁻³y = 1.37769× 10⁻³

y = 0.158 g and z = 0.156 Thus the mass of NaCl = 0.156 g and the mass percentage = 0.156/0.3146×100 = 49.7% NaCl

The mass percentage of NaCl(s) in the mixture is 49.7%