What is the resulting concentration when 250.0 mL of 0.65 M Magnesium Sulfate is added to 800.0 mL of water?
Assuming the volumes are additive (technically they are not)
0.65 x [250/(250+800)] = ?
0.15mol/liter
To find the resulting concentration when 250.0 mL of 0.65 M Magnesium Sulfate is added to 800.0 mL of water, we need to consider the dilution process.
Dilution Formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case:
C1 = 0.65 M (initial concentration of Magnesium Sulfate)
V1 = 250.0 mL (initial volume of Magnesium Sulfate)
C2 = ? (final concentration)
V2 = 250.0 mL + 800.0 mL = 1050.0 mL (final volume)
Now, let's substitute these values into the formula:
(0.65 M)(250.0 mL) = C2(1050.0 mL)
Solving for C2:
C2 = (0.65 M)(250.0 mL) / 1050.0 mL
Using the given values, the final concentration, C2, is approximately 0.154 M.
Therefore, the resulting concentration after adding 250.0 mL of 0.65 M Magnesium Sulfate to 800.0 mL of water is approximately 0.154 M.