Prepare 45.00ml of 0.125M gold III chloride solution, given 500.00ml of a 0.500M gold III chloride solution. Explain how you will prepare the solution
c1v1 = c2v2
0.500*x = 0.125*45
Solve for x = volume of the 0.500 M stuff you have. Pipet that amount into a 45 mL volumetric flask and make to the mark with DI water.
Yes, I know 45 mL volumetric flasks are not common. Neither are 11.25 mL pipets.
To prepare a 45.00 mL of 0.125 M gold III chloride solution, given a 500.00 mL of 0.500 M gold III chloride solution, you will need to use the equation C1V1 = C2V2, where C1 and V1 represent the initial concentration and volume, and C2 and V2 represent the final concentration and volume.
In this case, C1 is 0.500 M, V1 is 500.00 mL, C2 is 0.125 M, and V2 is 45.00 mL.
Now, let's solve the equation step by step:
Step 1: Calculate the amount of moles in the initial solution.
- Moles (n) = Concentration (C) x Volume (V)
- Moles (n1) = 0.500 M x 500.00 mL
- Note: 0.500 M represents 0.500 moles of gold III chloride dissolved in 1 liter of solution.
- Converting mL to L, Moles (n1) = 0.500 M x 0.500 L (500.00 mL = 0.500 L)
- Moles (n1) = 0.250 moles
Step 2: Calculate the volume (V2) needed for the desired concentration.
- Rearranging the equation C1V1 = C2V2, we get V2 = (C1V1) / C2
- V2 = (0.500 M x 500.00 mL) / 0.125 M
- V2 = 2000 mL
- Converting mL to L, V2 = 2.000 L
Step 3: Calculate the volume of the initial solution needed.
- Rearranging the equation C1V1 = C2V2, we get V1 = (C2V2) / C1
- V1 = (0.125 M x 45.00 mL) / 0.500 M
- V1 = 11.25 mL
Therefore, to prepare a 45.00 mL of 0.125 M gold III chloride solution from a 500.00 mL of 0.500 M gold III chloride solution, you need to take 11.25 mL of the initial solution and dilute it with a solvent (such as water) to a final volume of 45.00 mL.