Calculate the mass in grams of KMnO4 that should be used to prepare 2.0 L of 0.25 M KMnO4 solution.
The correct answer and explanation is :
To calculate the mass of potassium permanganate (KMnO₄) required to prepare 2.0 L of a 0.25 M solution:
Step-by-step Calculation:
- Use the formula:
$$
\text{Mass (g)} = \text{Molarity (mol/L)} \times \text{Volume (L)} \times \text{Molar Mass (g/mol)}
$$
- Known values:
- Molarity = 0.25 mol/L
- Volume = 2.0 L
- Molar mass of KMnO₄ = 39.10 (K) + 54.94 (Mn) + 4 × 16.00 (O)
$$
= 39.10 + 54.94 + 64.00 = 158.04 \text{ g/mol}
$$
- Plug in the values:
$$
\text{Mass} = 0.25 \times 2.0 \times 158.04 = 79.02 \text{ g}
$$
✅ Final Answer: 79.02 grams of KMnO₄
Explanation (≈300 words):
To prepare a specific concentration of a solution, we use the concept of molarity, which expresses the number of moles of solute per liter of solution. Here, we are asked to prepare 2.0 liters of a 0.25 M potassium permanganate (KMnO₄) solution. The molarity (M) tells us that each liter of this solution should contain 0.25 moles of KMnO₄.
First, determine how many moles of KMnO₄ are needed in total:
$$
0.25 \, \text{mol/L} \times 2.0 \, \text{L} = 0.50 \, \text{mol}
$$
Next, we convert moles to grams using the molar mass of KMnO₄. The molar mass is calculated by summing the atomic masses of its constituent elements:
- Potassium (K) = 39.10 g/mol
- Manganese (Mn) = 54.94 g/mol
- Oxygen (O) = 16.00 g/mol × 4 = 64.00 g/mol
Adding these gives the molar mass:
$$
39.10 + 54.94 + 64.00 = 158.04 \, \text{g/mol}
$$
Now multiply the number of moles by the molar mass:
$$
0.50 \, \text{mol} \times 158.04 \, \text{g/mol} = 79.02 \, \text{g}
$$
This means you need 79.02 grams of KMnO₄ to make 2.0 liters of a 0.25 M solution. Accurate measurement is crucial in lab work to ensure proper chemical reactions, especially with KMnO₄, a strong oxidizer used in redox titrations and disinfection.