Question

A mixture of KBr and NaBr weighing 0.560 g was treated with aqueous Ag+ and all the bromide ion was recovered as 0.970 g of pure AgBr. What was the fraction by weight of KBr in the sample?

(K = 39, Br = 80, Ag = 108 Na =23)

Answer 1

Explanation:

please refer to pic for solution

Answer 2

Answer:

Explanation:

Weight of the total mixture = 0.560g (Given)

Recovery of bromide ion = 0.970 g (Given)

Let the mass of KBr be = X then

Mass of NaBr = 0.560 - X  

To find the total mass of Br -  

= AgBr × Br/AgBr = 0.970 × 79.90 / 187.77

= 0.42552 g Br  

1 mol of KBr gives 1 mole of bromide ions.

1 mol of NaBr gives 1 mole of bromide ions.

Thus, according to the equation -

n(KBr) + n(NaBr) = n(AgBr)  

= (x/119)+((0.56-x)/102.9)

=(0.97/187.8)  

= x(KBr) = 0.21 g