Describe how a 250.0 mL aqueous solution of 1.50 M MgCl2 is prepared from a stock solution of 6.00 M MgCl2.
ANSWER: Measure 62.5 mL of the stock solution into a container and dilute to 250.0 mL
so what i need explained is where did 62.5 mL come from?
M1V1=M2V2
where M1=1.50 M, V1=250.0 mL, and M2=6.00M
just divide the 250mL by 4 to get 62.5, that cuts the molarity by 4 as well
To understand where the value of 62.5 mL comes from, we need to analyze the concept of dilution and the relationship between the molarity, volume, and concentration of a solution.
In this case, you have a stock solution of 6.00 M MgCl2 and you want to prepare a 250.0 mL solution with a concentration of 1.50 M. Dilution involves adding a solvent (usually water) to decrease the concentration of the solute.
The formula for dilution is given by:
C1V1 = C2V2
Where:
C1 = initial concentration (stock solution)
V1 = initial volume (stock solution)
C2 = final concentration (desired concentration)
V2 = final volume (desired volume)
In this case:
C1 = 6.00 M (stock concentration)
V1 = ? (stock volume)
C2 = 1.50 M (desired concentration)
V2 = 250.0 mL (desired volume)
Now, let's substitute the known values into the formula and solve for V1:
6.00 M * V1 = 1.50 M * 250.0 mL
V1 = (1.50 M * 250.0 mL) / 6.00 M
V1 = 62.5 mL
So, to prepare a 250.0 mL solution with a concentration of 1.50 M MgCl2, you need to measure 62.5 mL of the 6.00 M stock solution and dilute it with water to a final volume of 250.0 mL.