How many grams of AgNO3 are required to prepare 250. mL of a 0.135 M solution?

• A. 5.72 g
• B. 1.15 g
• C. 3.42 g
• D. 9.10 g

Answer: (A)

Verified step by step guidance

1. Identify the given information: volume of solution (V) = 250. mL and molarity (M) = 0.135 M. Convert the volume from milliliters to liters because molarity is in moles per liter. Use the conversion: \(V(\text{L}) = \frac{V(\text{mL})}{1000}\).
2. Calculate the number of moles of AgNO\_3 needed using the molarity formula: \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\). Rearranged, this is \(\text{moles of solute} = M \times V(\text{L})\).
3. Find the molar mass of AgNO\_3 by adding the atomic masses of silver (Ag), nitrogen (N), and oxygen (O) atoms. Use the periodic table values: Ag = 107.87 g/mol, N = 14.01 g/mol, O = 16.00 g/mol. Calculate \(\text{molar mass} = 107.87 + 14.01 + 3 \times 16.00\).
4. Calculate the mass of AgNO\_3 required by multiplying the moles of AgNO\_3 by its molar mass: \(\text{mass} = \text{moles} \times \text{molar mass}\).
5. Express the final answer in grams, which gives the mass of AgNO\_3 needed to prepare 250 mL of a 0.135 M solution.