Sure! To complete the table for predicting the concentration of the solution as water is added or removed from the tank, we can use the dilution equation [tex]\(M_1 V_1 = M_2 V_2\)[/tex]. This equation implies that the product of the initial concentration and volume equals the product of the final concentration and volume.
Given the initial concentration [tex]\(M_1\)[/tex] and volume [tex]\(V_1\)[/tex]:
- [tex]\(M_1 = 1.0\, \text{mol/L}\)[/tex]
- [tex]\(V_1 = 1.0\, \text{L}\)[/tex]
We can find the new concentration [tex]\(M_2\)[/tex] for different volumes [tex]\(V_2\)[/tex].
[tex]\[
M_2 = \frac{M_1 \times V_1}{V_2}
\][/tex]
Using this equation, we can calculate [tex]\(M_2\)[/tex] for each given volume [tex]\(V_2\)[/tex]:
1. For [tex]\(V_2 = 1.0\, \text{L}\)[/tex]:
[tex]\[
M_2 = \frac{1.0\, \text{mol/L} \times 1.0\, \text{L}}{1.0\, \text{L}} = 1.0\, \text{mol/L}
\][/tex]
2. For [tex]\(V_2 = 0.80\, \text{L}\)[/tex]:
[tex]\[
M_2 = \frac{1.0\, \text{mol/L} \times 1.0\, \text{L}}{0.80\, \text{L}} = 1.25\, \text{mol/L}
\][/tex]
3. For [tex]\(V_2 = 0.60\, \text{L}\)[/tex]:
[tex]\[
M_2 = \frac{1.0\, \text{mol/L} \times 1.0\, \text{L}}{0.60\, \text{L}} = 1.67\, \text{mol/L}
\][/tex]
4. For [tex]\(V_2 = 0.50\, \text{L}\)[/tex]:
[tex]\[
M_2 = \frac{1.0\, \text{mol/L} \times 1.0\, \text{L}}{0.50\, \text{L}} = 2.0\, \text{mol/L}
\][/tex]
5. For [tex]\(V_2 = 0.40\, \text{L}\)[/tex]:
[tex]\[
M_2 = \frac{1.0\, \text{mol/L} \times 1.0\, \text{L}}{0.40\, \text{L}} = 2.5\, \text{mol/L}
\][/tex]
6. For [tex]\(V_2 = 0.20\, \text{L}\)[/tex]:
[tex]\[
M_2 = \frac{1.0\, \text{mol/L} \times 1.0\, \text{L}}{0.20\, \text{L}} = 5.0\, \text{mol/L}
\][/tex]
Hence, the completed table is:
[tex]\[
\begin{array}{|l|l|}
\hline
\text{Volume (L)} & \text{Concentration (mol/L)} \\
\hline
1.0 & 1.0 \\
\hline
0.80 & 1.25 \\
\hline
0.60 & 1.67 \\
\hline
0.50 & 2.0 \\
\hline
0.40 & 2.5 \\
\hline
0.20 & 5.0 \\
\hline
\end{array}
\][/tex]
This table provides the predicted concentrations of the solution for various volumes of the tank.
