The nitric acid was purchased as a solution that has 70% HNO3. That volume (mL) of this solution is necessary for preparing 250 mL
Solution 2.0 mol L-1? The density of HNO3 (aq) 70% is 1.42 g cm-3
What is the molarity of the initial HNO3? That's
1000 cc x 1.42 g/cc x 0.70 x (1/63) = approx 16 M but you need to do that more accurately.
Then mL1 x M1 = mL2 x M2
mL1 x 16 = mL2 x 3.0
mL1 = ? = mL of the initial 70% solution.
To find the volume of the 70% HNO3 solution needed to prepare 250 mL of a 2.0 mol L-1 solution, we need to use the concept of the molarity and the density of the solution.
First, let's calculate the number of moles of HNO3 needed to prepare the 250 mL of the 2.0 mol L-1 solution:
Molarity (M) = Moles (mol) / Volume (L)
Rearranging the equation, we have:
Moles (mol) = Molarity (M) x Volume (L)
Substituting the given values:
Moles (mol) = 2.0 mol L-1 x 0.250 L
Moles (mol) = 0.5 mol
Since the 70% HNO3 solution is expressed in terms of its weight percentage, we need to find the weight of HNO3 required to obtain 0.5 mol.
The molar mass of HNO3 is:
1(H) + 14(N) + 3(16)(O) = 63 g/mol
Thus, the weight of HNO3 required is:
Weight (g) = Moles (mol) x Molar Mass (g/mol)
Weight (g) = 0.5 mol x 63 g/mol
Weight (g) = 31.5 g
Now, to find the volume of the 70% HNO3 solution needed, we can use the density of the solution.
Density (g/cm³) = Mass (g) / Volume (cm³)
Rearranging the equation, we have:
Volume (cm³) = Mass (g) / Density (g/cm³)
Substituting the given values:
Volume (cm³) = 31.5 g / 1.42 g/cm³
Volume (cm³) = 22.18 cm³
Since the density is given in grams per cubic centimeter (g/cm³), and we need the volume in milliliters (mL), we can convert it:
Volume (mL) = Volume (cm³) * 1 mL/cm³
Volume (mL) = 22.18 cm³ * 1 mL/cm³
Volume (mL) = 22.18 mL
Therefore, you would need 22.18 mL of the 70% HNO3 solution to prepare 250 mL of a 2.0 mol L-1 solution.