Final answer:
To prepare 100 mL of a 0.25 M oxalate (C2O4²-) solution, 3.55 grams of ammonium oxalate monohydrate ((NH₄)2C2O4⋅H2O) are needed.
Step-by-step explanation:
To calculate the number of grams of ammonium oxalate monohydrate (NH₄)2C2O4⋅H2O needed to prepare 100 mL of a 0.25 M oxalate (C2O4²-) anion solution, we must first calculate the moles of oxalate required using the molarity equation:
Molarity (M) = moles of solute / volume of solution in liters (L)
So, to find the moles of oxalate required for 100 mL (0.100 L) of a 0.25 M solution:
Moles = Molarity × Volume = 0.25 mol/L × 0.100 L = 0.025 moles
Since one mole of ammonium oxalate monohydrate yields one mole of oxalate ions upon dissociation, we need 0.025 moles of the salt. Next, we calculate the mass of the salt using its molar mass (MW = 142.11 g/mol for (NH₄)2C2O4⋅H2O):
Mass = moles × molar mass = 0.025 moles × 142.11 g/mol = 3.55275 g
Therefore, 3.55 grams of ammonium oxalate monohydrate are required.