Molarity calculation: How to prepare Stock Solution ? With example

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