how would you prepare 175.0ml of .150 m agno3 solution starting with pure agno3

How many moles do you want? That's M x L = ?

Then mols AgNO3 = grams AgNO3/molar mass AgNO3.

To prepare a 175.0 mL of 0.150 M AgNO3 solution starting with pure AgNO3, you will need to calculate the mass of AgNO3 required and then dissolve it in water to make the desired volume.

The formula to calculate the amount of solute required (in this case, AgNO3) is:

Mass of Solute (g) = Volume of Solution (L) × Concentration of Solution (M) × Molar Mass of Solute (g/mol)

Let's calculate the amount of AgNO3 needed:

1. Molar Mass of AgNO3:
The molar mass of AgNO3 is calculated by adding the atomic masses of its constituent elements:
AgNO3 = (1 × Ag) + (1 × N) + (3 × O)
AgNO3 = 107.87 g/mol + 14.01 g/mol + (3 × 16.00 g/mol)
AgNO3 = 169.87 g/mol

2. Mass of AgNO3:
Mass of AgNO3 = (175.0 mL ÷ 1000 mL/L) × (0.150 mol/L) × (169.87 g/mol)
Mass of AgNO3 = 4.245 g

Therefore, you need 4.245 grams of pure AgNO3 to prepare a 175.0 mL of 0.150 M AgNO3 solution.

To make the solution, weigh 4.245 grams of AgNO3 using a balance. Then, dissolve it in enough water to bring the total volume to 175.0 mL. Ensure the AgNO3 is fully dissolved before using the solution.

Note: When working with hazardous chemicals like AgNO3, it is essential to follow proper safety precautions, such as wearing protective gloves, goggles, and a lab coat, and working in a well-ventilated area.