Answer:
we need 1.605 grams of AgNO3 to prepare 237.3 mL of 0.0312 M solution.
Step-by-step explanation:
First, let's write down the formula for calculating the mass of a solute:
mass = molarity x volume x molar mass
We are given the volume and molarity of the solution, so we just need to find the molar mass of AgNO3.
AgNO3 has a molar mass of 169.87 g/mol.
Now we can substitute into the formula:
mass = 0.0312 mol/L x 0.2373 L x 169.87 g/mol
mass = 1.605 g
Therefore, we need 1.605 grams of AgNO3 to prepare 237.3 mL of 0.0312 M solution.