Calculate the amount (in grams) of AgNO3 needed to prepare 250.0 mL of 0.25 M of AgNO3.
M = mols/L
Solve for mols.
then mols = grams/molar mass. Solve for grams. Post your work if you get stuck.
To calculate the amount of AgNO3 needed to prepare a solution of a given concentration, we can use the formula:
amount (in grams) = concentration (in moles per liter) * volume (in liters) * molar mass (in grams per mole)
Given:
- Volume of solution = 250.0 mL = 0.2500 L
- Concentration of AgNO3 = 0.25 M (moles per liter)
To find the molar mass of AgNO3:
- Ag (Silver) has a molar mass of 107.87 g/mol
- N (Nitrogen) has a molar mass of 14.01 g/mol
- O (Oxygen) has a molar mass of 16.00 g/mol
Adding these together:
molar mass of AgNO3 = (1 * molar mass of Ag) + (1 * molar mass of N) + (3 * molar mass of O)
= (1 * 107.87) + (1 * 14.01) + (3 * 16.00)
= 107.87 + 14.01 + 48.00
= 169.88 g/mol
Now, let's substitute the given values into the formula to calculate the amount of AgNO3 needed:
amount (in grams) = 0.25 M * 0.2500 L * 169.88 g/mol
Performing the calculation:
amount (in grams) = 0.0625 mol * 169.88 g/mol
= 10.61 grams
Therefore, you would need approximately 10.61 grams of AgNO3 to prepare 250.0 mL of 0.25 M AgNO3 solution.