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.