How many grams of silver nitrate are needed to prepare 250 ml of a 0.100 M silver nitrate solution?

To calculate the amount of silver nitrate (AgNO₃) needed to prepare a 0.100 M solution in 250 ml, we first need to understand the definition of molarity (M). Molarity is defined as the number of moles of solute per liter of solution.

1. **Calculate the number of moles required**: Since the target concentration is 0.100 M, we first convert 250 ml to liters:

250 ml = 0.250 L

Now, using the molarity formula, we can find the number of moles of silver nitrate required:

Number of moles = Molarity × Volume = 0.100 moles/L × 0.250 L = 0.025 moles

2. **Calculate the mass of silver nitrate**: Next, we need to convert moles of silver nitrate to grams. To do this, we need the molar mass of silver nitrate. The molar mass can be calculated as follows:

Molar mass of AgNO₃ = Atomic mass of Ag + Atomic mass of N + 3 × Atomic mass of O

Atomic mass of Ag = 107.87 g/mol

Atomic mass of N = 14.01 g/mol

Atomic mass of O = 16.00 g/mol

Molar mass of AgNO₃ = 107.87 + 14.01 + (3 × 16.00) = 169.87 g/mol

Now, we multiply the number of moles by the molar mass to find the mass:

Mass = Number of moles × Molar mass = 0.025 moles × 169.87 g/mol = 4.247 g

Therefore, you will need approximately 4.25 grams of silver nitrate to prepare 250 ml of a 0.100 M silver nitrate solution.