How many grams of (NH_4)_3PO_4 are required to prepare 0.250 L of a 0.150 M (NH_4)_3PO_4 solution?

• A. 3.30 g
• B. 9.90 g
• C. 12.4 g
• D. 6.60 g

Answer: (B)

Verified step by step guidance

1. Identify the given information: volume of solution (V) = 0.250 L, molarity (M) = 0.150 M, and the solute is ammonium phosphate, (NH_4)_3PO_4.
2. Recall the definition of molarity: \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\). Use this to find the moles of (NH_4)_3PO_4 needed: \(\text{moles} = M \times V\).
3. Calculate the molar mass of (NH_4)_3PO_4 by summing the atomic masses of all atoms in the formula: 3 ammonium ions (NH_4)^+, 1 phosphate ion (PO_4)^{3-}.
4. Use the molar mass to convert moles of (NH_4)_3PO_4 to grams: \(\text{mass} = \text{moles} \times \text{molar mass}\).
5. The resulting mass is the amount of (NH_4)_3PO_4 required to prepare 0.250 L of a 0.150 M solution.