If I have 275 mL of a 1.5 M NaBr solution, what will the concentration be if I add 450 mL more water to it?
If dilute 580 mL of 1.0 M lithium acetate solution to a volume of 1250 mL, what will the concentration of this solution be?
Technically you can't calculate this because, technically the volumes are not additive. In practice they are with very little error in this case so assuming the volume are additive do it this way. Follow the same procedure for the second problem you posted.
mL1 x M1 = mL2 x M2
275 x 1.5 = 725 x M2
Solve for M2
To calculate the final concentration of the solution after adding more water, we need to consider the fact that the amount of solute (NaBr) remains the same while the volume of the solution increases.
First, let's calculate the initial amount of NaBr in the solution.
Given:
Volume (V1) = 275 mL
Concentration (C1) = 1.5 M
The initial amount of NaBr (A1) can be calculated using the formula:
A1 = V1 * C1
Substituting the given values:
A1 = 275 mL * 1.5 M
Next, we need to calculate the final volume of the solution by adding 450 mL of water to the initial solution.
Final Volume (V2) = V1 + Volume of water added
Substituting the given values:
V2 = 275 mL + 450 mL
Now, we can calculate the final concentration (C2) using the formula:
C2 = A1 / V2
Substituting the values we calculated earlier:
C2 = (275 mL * 1.5 M) / (275 mL + 450 mL)
Simplifying the equation:
C2 = 412.5 mL·M / 725 mL
C2 ≈ 0.568 M
Therefore, the final concentration of the NaBr solution, after adding 450 mL of water, will be approximately 0.568 M.