What is the molarity of a nitric acid solution prepared by diluting 250.0 mL of 6.00 M HNO3 to a total volume of 2.50 L?
you diluted it 10 times. Concentration: .6M
To find the molarity of a nitric acid solution prepared by dilution, we can use the formula:
M₁V₁ = M₂V₂
Where:
M₁ = initial molarity of the solution
V₁ = initial volume of the solution
M₂ = final molarity of the solution
V₂ = final volume of the solution
Given:
M₁ = 6.00 M (initial molarity)
V₁ = 250.0 mL (initial volume)
M₂ = ?
V₂ = 2.50 L (final volume)
First, we need to convert the initial volume from milliliters (mL) to liters (L):
V₁ = 250.0 mL = 0.250 L
Now, we can substitute the values into the formula and solve for M₂:
M₁V₁ = M₂V₂
(6.00 M)(0.250 L) = M₂(2.50 L)
Simplifying this equation:
1.50 mol = 2.50 L × M₂
Now, solve for M₂ by dividing both sides of the equation by 2.50 L:
M₂ = 1.50 mol / 2.50 L
M₂ = 0.60 M
Therefore, the molarity of the nitric acid solution prepared by diluting 250.0 mL of 6.00 M HNO3 to a total volume of 2.50 L is 0.60 M.