Answer :
Final answer:
To prepare 4.00 L of 0.175 M HClO4, the student needs to dilute 87.5 mL of the 8.00 M HClO4 solution with water until the volume reaches 4.00 L.
Explanation:
The preparation of a 0.175 M solution of HClO4 starting with an 8.00 M solution can be done by dilution. The formula for dilution is M1V1 = M2V2 where M represents molarity (moles/L) and V is volume. In this case, M1 is the initial molarity (8.00 M), M2 is the final molarity (0.175 M), and V2 is the final volume (4.00 L). To calculate the starting volume (V1), we can rearrange the equation to V1 = M2V2 / M1. By inputting the known quantities, we find out that V1 = (0.175 moles/L * 4.00 L) / 8.00 moles/L = 0.0875 L or 87.5 mL. So, to prepare 4.00 L of 0.175 M HClO4, you would dilute 87.5 mL of the 8.00 M solution with water until the volume reaches 4.00 L in the volumetric flask.
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Final answer:
To prepare 4.00 L of 0.175 M HClO4 from an 8.00 M stock solution, you need to dilute 0.875 L (875 mL) of the stock solution with water.
Explanation:
To prepare 4.00 L of 0.175 M HClO4 from an 8.00 M stock solution, you can use the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, the initial concentration (C1) is 8.00 M, the final concentration (C2) is 0.175 M, and the final volume (V2) is 4.00 L.
By rearranging the formula, we can solve for V1, which represents the volume of the stock solution to be diluted:
C1V1 = C2V2
(8.00 M)(V1) = (0.175 M)(4.00 L)
Dividing both sides of the equation by 8.00 M:
V1 = (0.175 M)(4.00 L) / 8.00 M
Simplifying the equation:
V1 = 0.875 L
Therefore, you need to dilute 0.875 L (875 mL) of the 8.00 M stock solution with water to obtain 4.00 L of 0.175 M HClO4 in a volumetric flask.
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