What is the final concentration (in M) of 17.6 mL sodium chloride solution with an initial concentration of 1.86 M which is diluted with 18.9 mL of solvent? Assume the volumes are additive.
1.86M x [17.6 mL/(17.6+18.9)]
To calculate the final concentration (in Molar, or M) of a solution after dilution, you can use the dilution equation:
C1V1 = C2V2
Where:
C1 is the initial concentration of the solution
V1 is the initial volume of the solution
C2 is the final concentration of the solution
V2 is the final volume of the solution
In this case, the initial volume (V1) is 17.6 mL, the initial concentration (C1) is 1.86 M, and the volume of solvent added (V2) is 18.9 mL.
First, let's convert the volumes to liters:
V1 = 17.6 mL = 17.6/1000 L = 0.0176 L
V2 = 18.9 mL = 18.9/1000 L = 0.0189 L
Now, let's substitute these values into the dilution equation and solve for the final concentration (C2):
C1V1 = C2V2
(1.86 M)(0.0176 L) = C2(0.0189 L)
Simplifying the equation:
0.032736 = 0.0189C2
Now, divide both sides of the equation by 0.0189 to solve for C2:
C2 = 0.032736 / 0.0189
C2 ≈ 1.731 M
Therefore, the final concentration is approximately 1.731 M.