Question

in an experiment, na2co3(aq) was added to 25.0 ml of cacl₂ solution until no more precipitate was formed. the mass of precipitate produced was 2.50 g. the concentration of the cacl₂ solution is​

Answer 1

Solution stoichiometry calculations involve chemical reactions taking place in solution.

Of the various methods of expressing solution concentration the most convenient for general

laboratory use is molarity, which is defined:

Moles of solute nsolute

Molarity = or M =

Liters of solution Lsoln

Chemical reactions are written in terms of moles of reactants and products; this molarity

concentration unit relates moles of solute to volume of solution. Thus, easily measured solution

volumes provide a simple method of measuring moles of reactants.

EXAMPLE: What is the molarity of a solution made by dissolving 5.67 g of potassium chloride

in enough water to make 100.0 mL of solution?

This data gives a relationship between amount of solute and volume of solution: 5.67 g KCl /

100.0 mL. To find molarity we must convert grams KCl to moles KCl and mL solution to L:

5.67 g KCl 1 mol KCl 1000 mL 0.760 mol KCl

x x = or 0.760 M KCl

100.0 mL 74.6 g KCl L L

Whenever the solution concentration is given in molarity, M, you must change to the equivalent

units, mol/L or mol/1000 mL, to use as a conversion factor.

EXAMPLE: What mass of solute is contained in 15.0 mL of a 0.760 M KCl solution?

The conversion sequence is:

mL solution → L solution → mol KCl → g KCl

1 L 0.760 mol KCl 74.6 g KCl

15.0 mL x x x = 0.850 g KCl

1000 mL L mol KCl