What volume for 1.5M KCl solution must be used to prepare 500. mL of a .550 M KCl solution?
I need the set up
500 ml *(.55 M/1.5 M)
=? ml
To determine the volume of the 1.5M KCl solution needed to prepare 500 mL of a 0.550 M KCl solution, we can use the equation:
C1V1 = C2V2
Where:
C1 refers to the concentration of the original solution
V1 refers to the volume of the original solution
C2 refers to the concentration of the final solution
V2 refers to the volume of the final solution
Let's plug in the given values:
C1 = 1.5M
V1 is what we need to find
C2 = 0.550M
V2 = 500 mL
Rearranging the equation, we have:
V1 = (C2 * V2) / C1
Substituting the values, we get:
V1 = (0.550M * 500 mL) / 1.5M
Now, we can calculate the volume:
V1 = (0.275 mol/L * 0.500 L) / 1.5 mol/L
V1 = 0.1375 L
Therefore, the volume of the 1.5M KCl solution needed to prepare 500 mL of a 0.550 M KCl solution is 0.1375 liters (or 137.5 mL).