Show calculations for how you will prepare 250 mL of 0.1 M AgNO3 solution from solid AgNO3 and water Give the reagent amounts and units.
How many moles do you want? That's M x L = ? mols.
Then mols = grams/molar mass,
You know mols from above and molar mass, solve for grams.
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To prepare a solution with a specific concentration (0.1 M) from a solid reagent (AgNO3) and water, we need to know the molar mass of AgNO3 and the desired volume of the solution.
1. Find the molar mass of AgNO3:
The molar mass can be calculated by adding up the atomic masses of each element in the compound.
Ag (Silver) has an atomic mass of approximately 107.87 g/mol,
N (Nitrogen) has an atomic mass of approximately 14.01 g/mol, and
O (Oxygen) has an atomic mass of approximately 16.00 g/mol.
Molar mass of AgNO3 = (1 x Ag) + (1 x N) + (3 x O)
Molar mass of AgNO3 = (1 x 107.87 g/mol) + (1 x 14.01 g/mol) + (3 x 16.00 g/mol)
Molar mass of AgNO3 = 169.87 g/mol
2. Calculate the amount of AgNO3 needed:
The molar concentration (M) is defined as the number of moles of solute per liter of solution.
So, to calculate the amount of AgNO3 needed, we use the formula:
moles = Molarity x Volume
Given:
Molarity (M) = 0.1 M
Volume (V) = 250 mL = 0.250 L
Moles of AgNO3 = 0.1 mol/L x 0.250 L
Moles of AgNO3 = 0.025 mol
Now, we need to convert the moles of AgNO3 to grams using the molar mass:
Mass of AgNO3 = Moles of AgNO3 x Molar mass of AgNO3
Mass of AgNO3 = 0.025 mol x 169.87 g/mol
Mass of AgNO3 = 4.2475 g
Therefore, to prepare 250 mL of a 0.1 M AgNO3 solution, you will need approximately 4.25 grams of solid AgNO3 and enough water to make up the volume to 250 mL.