0.274 g manganese chlorate (Mn(ClO3)2) is dissolved in water to make 100.0 mL of solution. Calculate the molarity of manganese chlorate in solution. (b) Calculate the mass percent of manganese chlorate in a solution made by dissolving 0.274 g Mn(ClO3)2 in 10.00 g water.
The Correct Answer and Explanation is:
(a) Molarity of manganese chlorate (Mn(ClO₃)₂)
Step 1: Calculate the molar mass of Mn(ClO₃)₂.
- The molar mass of Mn (Manganese) = 54.94 g/mol
- The molar mass of Cl (Chlorine) = 35.45 g/mol
- The molar mass of O (Oxygen) = 16.00 g/mol
For Mn(ClO₃)₂:
- The molar mass of Mn(ClO₃)₂ = 54.94 g/mol + 2 × (35.45 g/mol + 3 × 16.00 g/mol)
= 54.94 g/mol + 2 × (35.45 g/mol + 48.00 g/mol)
= 54.94 g/mol + 2 × 83.45 g/mol
= 54.94 g/mol + 166.90 g/mol
= 221.84 g/mol
Step 2: Calculate the moles of Mn(ClO₃)₂.
- Mass of Mn(ClO₃)₂ = 0.274 g
- Moles of Mn(ClO₃)₂ = mass / molar mass
Moles of Mn(ClO₃)₂ = 0.274 g / 221.84 g/mol
Moles of Mn(ClO₃)₂ = 1.236 × 10⁻⁴ mol
Step 3: Calculate the molarity.
- Volume of solution = 100.0 mL = 0.1000 L
- Molarity (M) = moles of solute / volume of solution (in liters)
Molarity (M) = 1.236 × 10⁻⁴ mol / 0.1000 L
Molarity (M) = 1.236 × 10⁻³ M
(b) Mass Percent of Mn(ClO₃)₂ in solution
Step 1: Calculate the total mass of the solution.
- Mass of Mn(ClO₃)₂ = 0.274 g
- Mass of water = 10.00 g
Total mass of solution = 0.274 g + 10.00 g = 10.274 g
Step 2: Calculate the mass percent.
Mass percent = (mass of solute / total mass of solution) × 100%
Mass percent = (0.274 g / 10.274 g) × 100%
Mass percent ≈ 2.67%
Final Answers:
(a) The molarity of manganese chlorate in solution is 1.236 × 10⁻³ M.
(b) The mass percent of manganese chlorate in the solution is 2.67%.
This method ensures precise and reliable calculations, using basic principles of molar mass, moles, and mass percent.
