Concentration and Molarity Worked Example Problem

Preparing a Stock Solution

It's common to prepare stock solutions for chemistry labs.
It's common to prepare stock solutions for chemistry labs. Compassionate Eye Foundation/Tom Grill, Getty Images


a) Explain how to prepare 25 liters of a 0.10 M BaCl2 solution, starting with solid BaCl2.
b) Specify the volume of the solution in (a) needed to get 0.020 mol of BaCl2.


Part a): Molarity is an expression of the moles of solute per liter of solution, which can be written:

molarity (M) = moles solute / liters solution

Solve this equation for moles solute:

moles solute = molarity × liters solution

Enter the values for this problem:

moles BaCl2 = 0.10 mol/liter &times 25 liter
moles BaCl2 = 2.5 mol

To determine how many grams of BaCl2 are needed, calculate the weight per mole. Look up the atomic masses for the elements in BaCl2 from the Periodic Table. The atomic masses are found to be:

Ba = 137
Cl = 35.5

Using these values:

1 mol BaCl2 weighs 137 g + 2(35.5 g) = 208 g

So the mass of BaCl2 in 2.5 mol is:

mass of 2.5 moles of BaCl2 = 2.5 mol × 208 g / 1 mol
mass of 2.5 moles of BaCl2 = 520 g

To make the solution, weigh out 520 g of BaCl2 and add water to get 25 liters.

Part b): Rearrange the equation for molarity to get:

liters of solution = moles solute / molarity

In this case:

liters solution = moles BaCl2 / molarity BaCl2
liters solution = 0.020 mol / 0.10 mol/liter
liters solution = 0.20 liter or 200 cm3


Part a). Weigh out 520 g of BaCl2. Stir in sufficient water to give a final volume of 25 liters.

Part b). 0.20 liter or 200 cm3