How many ml of a 0.300M AgNO3 solution Will it take to make 500ml of a 0.100 M AgNO3 solution?

To determine how many milliliters of a 0.300 M AgNO3 solution are needed to prepare 500 mL of a 0.100 M AgNO3 solution, we can use the concept of dilution, which is based on the principle that the number of moles of solute before dilution equals the number of moles after dilution. This relationship can be expressed with the formula:

Understanding the Dilution Formula

The dilution formula is given by:

C1V1 = C2V2

Where:

• C1 = initial concentration of the solution (0.300 M)
• V1 = volume of the initial solution we need to find
• C2 = final concentration of the diluted solution (0.100 M)
• V2 = final volume of the diluted solution (500 mL)

Plugging in the Values

Now, we can substitute the known values into the equation:

0.300 M × V1 = 0.100 M × 500 mL

Solving for V1

To find V1, we rearrange the equation:

V1 = (0.100 M × 500 mL) / 0.300 M

Calculating this gives:

V1 = (50 mL) / 0.300

V1 = 166.67 mL

Final Calculation and Interpretation

This means you will need approximately 166.67 mL of the 0.300 M AgNO3 solution to create 500 mL of a 0.100 M AgNO3 solution. It's important to measure accurately and ensure that you mix the solutions thoroughly to achieve the desired concentration.

Practical Application

In a laboratory setting, always remember to add the concentrated solution to water, not the other way around, to ensure safety and proper mixing. This method of dilution is widely used in chemistry for preparing solutions of desired concentrations from stock solutions.