Prepare 250mls of 0.1N AgNO3 stock solution. How much gram of AgNO3 is needed to prepare 250mls of 0.1N AgNO3?

To prepare 250mls of 0.1N AgNO3 stock solution, we need to calculate the amount of AgNO3 needed.

First, we need to calculate the molecular weight of AgNO3:
Ag = 107.87 g/mol
N = 14.01 g/mol
O = 16.00 g/mol

Molecular weight of AgNO3 = 107.87 + 14.01 + (3 x 16.00) = 169.87 g/mol

Now, we can calculate the amount of AgNO3 needed to prepare 250mls of 0.1N solution using the formula:

Amount (g) = Normality (N) x Equivalent weight (g/mol) x Volume (L)

Amount (g) = 0.1 x 169.87 x 0.25 = 4.24675g

Therefore, we need 4.24675g of AgNO3 to prepare 250mls of 0.1N AgNO3 stock solution.

Step 1: Calculate the number of moles of AgNO3 needed

Molarity (N) = Equivalents/Liter
0.1 N = 0.1 eq/L

1 equivalent of AgNO3 = 1 mole of Ag+
Therefore, 0.1 eq/L = 0.1 mol/L

The volume of solution needed is 250 mL, which is 0.25 L.

Moles of AgNO3 = Concentration × Volume
Moles of AgNO3 = 0.1 mol/L × 0.25 L
Moles of AgNO3 = 0.025 mol

Step 2: Calculate the mass of AgNO3 needed

Molar mass of AgNO3 = 169.87 g/mol

Mass of AgNO3 = Moles of AgNO3 × Molar mass
Mass of AgNO3 = 0.025 mol × 169.87 g/mol
Mass of AgNO3 = 4.247 grams

Therefore, 4.247 grams of AgNO3 is needed to prepare 250 mL of 0.1 N AgNO3 stock solution.

Based on the first question what solution preparation method was used?