How many grams of CuSO4·5H2O must you weigh out in order to prepare 100.0 mL of a 0.10 M CuSO4 solution? (Molar mass of CuSO4·5H2O is 250 g·mol–1).

Question

BlackTom AI · Accepted Answer

To prepare a 0.10 M solution in 100.0 mL, first convert the volume to liters: 100.0 mL = 0.100 L. Then calculate the number of moles required using M = n/V, so n = M × V = 0.10 mol/L × 0.100 L = 0.0100 mol.
Next, use the molar mass to find the mass of CuSO4·5H2O needed: mass = moles × molar mass = 0.0100 mol × 250 g/mol = 2.50 g.
Now evaluate the options:
- 2.5 g: This matches the calculated mass exactly, so this option is correct. The slight difference in formatting (2.5 vs 2.50) is due to significant figures, but both represent the same mass.
- 250 g: This is 100 times larger than the correct mass. It would imply 1.00 mol of CuSO4·5H2O in 0.100 L, giving a 10.0 M solution, not 0.10 M, which is inconsistent with the target concentration.
- 0.25 g: This corresponds to 0.0010 mol (0.25 g ÷ 250 g/mol), which would yield a 0.010 M solution when diluted to 0.100 L, not the required 0.10 M; this shows a miscalculation by a factor of 10.
- 25 g: This equals 0.100 mol (25 g ÷ 250 g/mol), producing a 1.0 M solution in 0.100 L, again far from the intended 0.10 M.
In summary, the only option that aligns with both the molarity and volume is 2.5 g, since it yields the necessary 0.0100 mol of CuSO4·5H2O for 0.100 L at 0.10 M.

How many grams of CuSO4·5H2O must you weigh out in order to prepare 100.0 mL of a 0.10 M CuSO4 solution? (Molar mass of CuSO4·5H2O is 250 g·mol–1).单项选择题

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