calculate the final concentration if 4.0 L of 1.5 M NaC1 and 2.0 L of 3.0 M NaC1 are mixed.
Find moles NaCl from 1.5 M and moles NaCl from 3.0 M, add them together and divide by the total volume in liters. Post your work if you get stuck.
Moles of salt present: 4*1.5+2*3=12 moles salt.
Volume= 6liters
Molarity=moles/volume
To calculate the final concentration of a solution when two solutions are mixed, you need to use the formula for dilution:
C1V1 + C2V2 = CfVf
Where:
C1 = initial concentration of the first solution
V1 = volume of the first solution
C2 = initial concentration of the second solution
V2 = volume of the second solution
Cf = final concentration
Vf = final volume
In this case, the initial concentration of the first solution (1.5 M NaCl) is C1 = 1.5 M, and the volume of the first solution is V1 = 4.0 L. The initial concentration of the second solution (3.0 M NaCl) is C2 = 3.0 M, and the volume of the second solution is V2 = 2.0 L.
Let's substitute these values into the dilution formula:
(1.5 M) * (4.0 L) + (3.0 M) * (2.0 L) = Cf * Vf
6.0 mol + 6.0 mol = Cf * Vf
12.0 mol = Cf * Vf
Now, let's assume the final volume will be the sum of the volumes of the two solutions:
Vf = V1 + V2
Vf = 4.0 L + 2.0 L
Vf = 6.0 L
Substituting this value back into the equation:
12.0 mol = Cf * 6.0 L
To find Cf, we can rearrange the equation:
Cf = 12.0 mol / 6.0 L
Cf = 2.0 M
Therefore, the final concentration of NaCl in the mixture is 2.0 M.