Concentration and Molarity Worked Example Problem

Preparing a Stock Solution

By Anne Marie Helmenstine, Ph.D.,

Question: 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.
Solution:
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 × 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
Answer
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