**Question:**

a) Explain how to prepare 25 liters of a 0.10 M BaCl_{2} solution, starting with solid BaCl_{2}.

b) Specify the volume of the solution in (a) needed to get 0.020 mol of BaCl_{2}.

**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 BaCl_{2} = 0.10 mol/liter × 25 liter

moles BaCl_{2} = 2.5 mol

To determine how many grams of BaCl_{2} are needed, calculate the weight per mole. Look up the atomic masses for the elements in BaCl_{2} from the Periodic Table. The atomic masses are found to be:

Ba = 137

Cl = 35.5

Using these values:

1 mol BaCl_{2} weighs 137 g + 2(35.5 g) = 208 g

So the mass of BaCl_{2} in 2.5 mol is:

mass of 2.5 moles of BaCl_{2} = 2.5 mol × 208 g / 1 mol

mass of 2.5 moles of BaCl_{2} = 520 g

To make the solution, weigh out 520 g of BaCl_{2} 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 BaCl_{2} / molarity BaCl_{2}

liters solution = 0.020 mol / 0.10 mol/liter

liters solution = 0.20 liter or 200 cm^{3}

**Answer**

Part a). Weigh out 520 g of BaCl_{2}. Stir in sufficient water to give a final volume of 25 liters.

Part b). 0.20 liter or 200 cm^{3}