A sample of commercial perchloric acid is 70.0% HClO4 by mass; its density is 1.664 g/mL. How many milliliters of this concentrated HClO4 would be required to prepare 500. mL of 1.50 M HClO4 solution?

The volume of the conc. HClO4 to prepare 500 ml of 1.50 M solution is 64.6 ml.

Based on the given information,

•The mass of HClO4 is 70% by mass, the density of HClO4 is 1.664 g/ml or 1664 g/L.

•It is known that the molar mass of perchloric acid is 100.465 g/mol.  

Now the mass of HClO4 is,

Now the moles can be calculated as,

Now the molarity is,

Now, M1 is 1.50 M, M2 is 11.596 M, V1 is 0.500 L. There is a need to find V2, which can be done by using the dilution law, that is,

Now putting the values we get,

Thus, 64.6 ml of the concentrated HClO4 is need to prepare 500 ml of 1.50 M HClO4 solution.

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We need 64.7 mL of this concentrated HClO4 solution

Explanation:

Step 1: Data given

Percentage of HClO4 solutions = 70.0 %

Density of HClO4 = 1.664 g/mL

Volume of prepared HClO4 solution  =  500.0 mL

Molarity of the prepared HClO solution = 1.50 M

Step 2: Calculate mass HClO4

Let's assume we have 1 L = 1000 mL of solution

Density = mass / volume

Mass = density * volume

Mass = 1.664 g/mL * 1000 mL

Mass = 1664 grams

For a 70.0 M HClO4 solution the mass HClO4 is:

1664 g * 0.70 = 1165 grams HClO4

Step 3: Calculate molarity

1165 g HClO4 / 100.5 g  = 11.59 moles HClO4 per L = 11.59 M  

Step 4: Calculate the volume needed

C1V1 = C2V2  

⇒with C1 = the molarity of the 70.0 % HClO4 = 11.59 M

⇒with V1 = the volume of the 70.0 % HClO4 = TO BE DETERMINED

⇒with C2 = the molarity of the prepared HClO4 solution = 1.50 M

⇒with V2 = the volume of the prepared HClO4 solution = 500.0 mL = 0.500 L

11.59 * V2 = 1.50 M * 0.500

V2 = (1.50 * 0.500) / 11.59

V2 = 0.0647 L = 64.7 mL

We need 64.7 mL of this concentrated HClO4 solution

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.