Calculate the mass of KMnO
(in g) required to make 414 mL of 0.02 M KMnO
solution. (Note: To make the concentration as accurate as possible, an analytical balance will be used to weigh KMnO
. The analytical balance can read to 0.0001 g. So, keep four decimal places in the answer.) Answer:
The Correct Answer and Explanation is:
To calculate the mass of KMnO₄ (potassium permanganate) required to make 414 mL of a 0.02 M solution, follow these steps:
Step 1: Convert mL to L
We need the volume in liters (L), because molarity (M) is moles per liter.414 mL=0.414 L414\, \text{mL} = 0.414\, \text{L}414mL=0.414L
Step 2: Use the molarity formula
Molarity (M) = moles of solute / liters of solution
Rearranged:moles of KMnO₄=M×V=0.02×0.414=0.00828 mol\text{moles of KMnO₄} = M \times V = 0.02 \times 0.414 = 0.00828\, \text{mol}moles of KMnO₄=M×V=0.02×0.414=0.00828mol
Step 3: Find the molar mass of KMnO₄
Use atomic masses from the periodic table:
- K = 39.10 g/mol
- Mn = 54.94 g/mol
- O = 16.00 g/mol × 4 = 64.00 g/mol
Molar mass of KMnO₄=39.10+54.94+64.00=158.04 g/mol\text{Molar mass of KMnO₄} = 39.10 + 54.94 + 64.00 = 158.04\, \text{g/mol}Molar mass of KMnO₄=39.10+54.94+64.00=158.04g/mol
Step 4: Calculate the mass
Now use the number of moles and multiply by the molar mass:mass=moles×molar mass=0.00828×158.04=1.3073 g\text{mass} = \text{moles} \times \text{molar mass} = 0.00828 \times 158.04 = 1.3073\, \text{g}mass=moles×molar mass=0.00828×158.04=1.3073g
Final Answer:
1.3073 g\boxed{1.3073\, \text{g}}1.3073g
Explanation
To prepare a chemical solution with precision, it is essential to use proper units and conversion steps. The molarity (M) represents the number of moles of solute per liter of solution. Here, we are given a volume of 414 milliliters, which needs to be converted to liters because the molarity unit includes liters, not milliliters.
After converting to 0.414 liters, we multiply this by the given molarity (0.02 M) to find the moles of KMnO₄ needed. This gives 0.00828 moles. Next, we calculate the molar mass of KMnO₄ by summing the atomic masses of potassium (K), manganese (Mn), and four oxygen (O) atoms. This totals 158.04 g/mol.
Multiplying the number of moles by the molar mass gives us the mass of KMnO₄ required for the solution. This results in 1.3073 grams.
Since an analytical balance that reads to 0.0001 grams is used, the answer is rounded to four decimal places to ensure measurement precision. This approach ensures accurate and reliable preparation of a chemical solution, which is critical in laboratory settings where small errors can impact experimental outcomes.
