Question

A 63−g sample of impure KClO3 (solubility = 7.1 g per 100 g H2O at 20°C) is contaminated with 19 percent of KCl (solubility = 25.5 g per 100 g of H2O at 20°C).

(a) Calculate the minimum quantity of 20°C water needed to dissolve all the KCl from the sample.

(b) How much KClO3 will be left after this treatment? (Assume that the solubilities are unaffected by the presence of the other compound.)

Answer 1

Solution :-

Part a) Mass of sample = 63 g with 19 % impurity of KCl

Mass of KCl in the sample = 63 g * 19 % / 100 % = 11.97 g

Solubility of the KCl = 25.5 g per 100 g water at 20 C

Lets calculate the mass of water needed to dissolve the 11.97 g KCl

11.97 g KCl * 100 g water / 25.5 g KCl = 46.94 g water

Therefore we need 46.94 g water

Part b)

Solubility of the KClO3 is 7.1 g in 100 g water at 20 C

So lets calculate the mass of KCl dissolved in 46.94 g water

46.94 g water * 7.1 g / 100 g water = 3.33 g KClO3

Now lets calculate the mass of KClO3 left

Mass of KClO3 left = mass of sample – ( mass of KCl + mass of KClO3 dissolved)

                                  = 63 g – ( 11.97 g + 3.33 g)

                                 = 47.7 g

Hence mass of KClO3 remain is 47.7 g