Respuesta :
Answer:
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
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,
[tex]Moles = \frac{Mass}{Molar mass}[/tex]
[tex]Moles = \frac{1165 g}{100.465 g/mol} \\Moles = 11.596 moles[/tex]
Now the molarity is,
[tex]Molarity = \frac{Moles}{Liter} \\Molarity = \frac{11.596}{1 L} \\Molarity = 11.596 M[/tex]
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,
[tex]M1V1 = M2V2[/tex]
Now putting the values we get,
[tex]V2 = \frac{M1V1}{M2} \\[/tex]
[tex]V2 = \frac{1.50 * 0.500}{11.596} \\[/tex]
[tex]V2 = 0.0646 L\\V2 = 64.6 ml[/tex]
Thus, 64.6 ml of the concentrated HClO4 is need to prepare 500 ml of 1.50 M HClO4 solution.
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