6.7 mL of a ptassium chlorid solution was added to a 54.730 gevaporating dish. The combination weight 61.945 g. Afterevaporation the dish contents weight 55.428 g.

a. What was the mass percent of potassium chloride in thesolution?

b. If the actual mass percent of the potassium in the abovesolution was 10.00%, what was the percentage error of the abovemeasurment?

c. Why was the evaporation used to determine the masspercentage of potassium cloride in the solution rather than thefiltration or recrystallization? Explain.

Question

Chemistry

Guest

30 November, 17:48

6.7 mL of a ptassium chlorid solution was added to a 54.730 gevaporating dish. The combination weight 61.945 g. Afterevaporation the dish contents weight 55.428 g.

a. What was the mass percent of potassium chloride in thesolution?

b. If the actual mass percent of the potassium in the abovesolution was 10.00%, what was the percentage error of the abovemeasurment?

c. Why was the evaporation used to determine the masspercentage of potassium cloride in the solution rather than thefiltration or recrystallization? Explain.

+1

Answers (1)

Know the Answer?

Kristina Keith · Answer 1

A.

Mass of evaporation dish + solution of KCl = mass of dish + mass of solution of KCl

61.945 = 54.73 + mass of KCl solution

= 7.215 g

Mass of evaporation dish + dry KCl = mass of dish + mass of dry KCl

55.428 = 54.73 + mass of dry KCl

= 0.698 g.

% Mass of KCl = mass of dry KCl/mass of KCl solution * 100

= 0.698/7.215 * 100

= 9.67%

B.

At 10%, mass = 0.7215 g

Change in error = 0.7215 - 0.698

= 0.0235 g

% error = change in error/actual mass at 10% * 100

= 0.0235/0.7215 * 100

= 3.26 %

C.

In evaporation, KCl does not decompose on heating. In filtration, there needs to be a heterogeneous solution (particles or suspension of KCl present) and it is not suitable for soluble salts while recrystallisation is mainly a purification process for solution and KCl is equally soluble in cold and hot water.