Questions: How many milliliters of the approximately 0.015 M copper(II) sulfate solution are needed to prepare 100.00 milliliters of a 0.00030 M copper(II) sulfate solution? mL

How many milliliters of the approximately 0.015 M copper(II) sulfate solution are needed to prepare 100.00 milliliters of a 0.00030 M copper(II) sulfate solution? mL
Transcript text: 1) How many milliliters of the approximately 0.015 M copper(II) sulfate solution are needed to prepare 100.00 milliliters of a 0.00030 M copper(II) sulfate solution? $\square$ mL
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Solution

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Solution Steps

Step 1: Understand the Dilution Concept

To solve this problem, we need to use the concept of dilution, which is described by the formula:

\[ C_1V_1 = C_2V_2 \]

where:

  • \(C_1\) is the initial concentration of the solution,
  • \(V_1\) is the volume of the initial solution needed,
  • \(C_2\) is the final concentration of the solution,
  • \(V_2\) is the final volume of the solution.
Step 2: Identify Given Values

From the problem, we have:

  • \(C_1 = 0.015 \, \text{M}\) (initial concentration),
  • \(C_2 = 0.00030 \, \text{M}\) (final concentration),
  • \(V_2 = 100.00 \, \text{mL}\) (final volume).

We need to find \(V_1\), the volume of the initial solution required.

Step 3: Apply the Dilution Formula

Substitute the given values into the dilution formula:

\[ 0.015 \times V_1 = 0.00030 \times 100.00 \]

Step 4: Solve for \(V_1\)

Calculate \(V_1\) by rearranging the equation:

\[ V_1 = \frac{0.00030 \times 100.00}{0.015} \]

\[ V_1 = \frac{0.0300}{0.015} \]

\[ V_1 = 2.000 \]

Final Answer

The volume of the approximately 0.015 M copper(II) sulfate solution needed is \(\boxed{2.000 \, \text{mL}}\).

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