What is the molarity of a solution prepared by diluting 25.0 mL of 0.412 M magnesium nitrate to a volume of 1.50 L
please help!
Use the dilution formula of
c1v1 = c2v2
c = concn
v = volume.
To find the molarity of the solution, we need to use the formula:
M1V1 = M2V2
Where:
M1 = initial molarity
V1 = initial volume
M2 = final molarity
V2 = final volume
In this case, the initial molarity (M1) is 0.412 M and the initial volume (V1) is 25.0 mL. We also know that the final volume (V2) is 1.50 L.
Now, let's plug in the values into the formula and solve for M2:
(0.412 M) × (25.0 mL) = M2 × (1.50 L)
To continue solving, we need to convert mL to L:
25.0 mL = 25.0 mL × (1 L / 1000 mL) = 0.025 L
Plugging in the values again:
(0.412 M) × (0.025 L) = M2 × (1.50 L)
0.0103 = M2 × 1.50
Now, we can solve for M2 by dividing both sides of the equation by 1.50:
0.0103 / 1.50 = M2
M2 ≈ 0.0069 M
Therefore, the molarity of the diluted solution is approximately 0.0069 M.