Question

A 260.mL sample of a 0.315M CaCl2 solution is left on a hot plate overnight; the following

morning the solution is 1.20M. What volume of water evaporated from the 0.315M CaCl2
solution?

Answer 1

To find the volume of water evaporated from the 0.315M CaCl2 solution, you can use the concept of dilution.

The formula for dilution is:

\(M_1V_1 = M_2V_2\)

Where:
- \(M_1\) is the initial molarity of the solution
- \(V_1\) is the initial volume of the solution
- \(M_2\) is the final molarity of the solution
- \(V_2\) is the final volume of the solution

In this case, you have:

\(M_1 = 0.315M\) (initial molarity)
\(M_2 = 1.20M\) (final molarity)
\(V_1 = 260 mL\) (initial volume)

You want to find \(V_2\) (final volume), which represents the volume of the solution after water has evaporated.

Using the formula, you can rearrange it to solve for \(V_2\):

\(V_2 = \frac{M_1V_1}{M_2}\)

\(V_2 = \frac{0.315M \times 260 mL}{1.20M}\)

Now, calculate \(V_2\):

\(V_2 = \frac{81.9 mL}{1.20}\)

\(V_2 \approx 68.25 mL\)

So, approximately 68.25 mL of water evaporated from the 0.315M CaCl2 solution.