Question

a 50-g sample of impure kclo3 (solubility = 7.1 g per 100 g h2o at at 20 degrees c) is contaminated with 10 percent of kcl (solubility = 25.5 g per 100 g of h2o at 20 degrees c). calculate the minimum quantity of 20 degrees c water needed to dissolve all the kcl from the sample. how much kclo3 will be left after this treatment? ( Please, with regard to the effect of two solutes in one solvent )

Answer 1

Given :

Impure sample of KCl = 50.0 g , solubility = 7.1 g / 100 g H2O at 20⁰C

KCl is 10 % .

Solution:

Step 1 :

Calculation of mass of KCl

Mass of KCl = 50.0 g KClO3 x 10 g / 100 g

=5.0 g

Calculation of mass of water required to dissolve 5.0 g KCl

= 5.0 g KCl x 100 g H2O / 25.5 g KCl )

= 19.61 g H2O.

Calculation of mass of KClO3

Mass of KClO3 = 50.0 g – 5.0 g = 45.0 g KClO3

Mass of KClO3 dissolved in 19.61 g H2O

= 19.61 g H2O x 7.1 g / 100 g H2O

= 1.40 g KClO3

So the mass of KClO3 undissolved

= total mass of KClO3 – Mass of KClO3 dissolved in H2O

= 45.0 g – 1.40 g = 43.6 g KClO3

So answer is : 43.6 g KClO3