A nitric acid solution at 25% by weight has a density of 1.21 KG / L Calculate the volume needed to be taken to obtain 2 L of a 2.5 M solution of nitric acid
To calculate the volume of the 25% nitric acid solution needed to obtain 2 L of a 2.5 M solution, we need to use the equation:
C1*V1 = C2*V2
Where:
C1 = concentration of the initial solution (25%)
V1 = volume of the initial solution (unknown)
C2 = concentration of the final solution (2.5 M)
V2 = volume of the final solution (2 L)
First, let's convert the concentration of the initial solution from weight percent to molarity using the density of the solution:
Concentration (molarity) = (Concentration (weight %) * density) / (molar mass of nitric acid)
The molar mass of nitric acid (HNO3) is 63 g/mol.
Concentration (molarity) = (25% * 1.21 KG/L) / (63 g/mol)
To convert from kilograms to grams, we multiply by 1000.
Concentration (molarity) = (25% * 1.21 KG/L * 1000 g/KG) / (63 g/mol)
Concentration (molarity) = 483 * 1000 / 63 M
Concentration (molarity) = 7674.60 M
Now we can substitute the values into the equation:
(7674.60 M) * V1 = (2.5 M) * (2 L)
Solving for V1:
V1 = (2.5 M * 2 L) / 7674.60 M
V1 ≈ 0.000651 L
So, approximately 0.000651 liters (or 0.651 milliliters) of the 25% nitric acid solution is needed to obtain 2 L of a 2.5 M solution.