Final answer:
The percent water in CuSO₄·5H₂O is calculated by dividing the total mass of water in the hydrate by the molar mass of the entire hydrate and then converting that to a percentage. The calculation yields a percent water of 36.1%. The correct option is b.
Step-by-step explanation:
To determine the percent water in CuSO₄·5H₂O, you must first calculate the molar masses of CuSO₄ and H₂O. The molar mass of CuSO₄ (copper(II) sulfate) is approximately 159.61 g/mol, and the molar mass of H₂O (water) is approximately 18.015 g/mol. Since there are five water molecules per formula unit of copper(II) sulfate pentahydrate, the total molar mass of the water in the compound is 5 × 18.015 g/mol = 90.075 g/mol.
Next, add the molar mass of CuSO₄ and the total molar mass of the water to find the molar mass of the hydrate: 159.61 g/mol (CuSO₄) + 90.075 g/mol (5H₂O) = 249.685 g/mol.
Then, calculate the percent water by dividing the total mass of water by the total mass of the hydrate and multiplying by 100 to get a percentage: (90.075 g/mol ÷ 249.685 g/mol) × 100 = 36.08%. This value rounds to 36.1%, which matches the option (b).