Sure, let's work through the solution step-by-step:
1. Convert the volume of the unknown liquid from liters to cubic centimeters:
[tex]\[
1 \text{ liter} = 1000 \text{ cubic centimeters (cm}^3\text{)}
\][/tex]
Given:
[tex]\[
\text{Volume}_{\text{unknown}} = 0.890 \text{ L}
\][/tex]
Therefore:
[tex]\[
\text{Volume}_{\text{unknown}} = 0.890 \text{ L} \times 1000 \text{ cm}^3/\text{L} = 890 \text{ cm}^3
\][/tex]
2. Calculate the density of the unknown liquid using the formula:
[tex]\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\][/tex]
Given:
[tex]\[
\text{Mass}_{\text{unknown}} = 831 \text{ g}
\][/tex]
So:
[tex]\[
\text{Density}_{\text{unknown}} = \frac{831 \text{ g}}{890 \text{ cm}^3} \approx 0.934 \text{ g/cm}^3
\][/tex]
3. Round the density to 3 significant digits:
[tex]\[
\text{Density}_{\text{unknown}} \approx 0.934 \text{ g/cm}^3
\][/tex]
4. Compare the calculated density to the known densities of possible liquids:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Liquid} & \text{Density (g/cm}^3\text{)} \\
\hline
\text{Acetone} & 0.79 \\
\hline
\text{Methyl Acetate} & 0.93 \\
\hline
\text{Pentane} & 0.63 \\
\hline
\text{Chloroform} & 1.5 \\
\hline
\text{Glycerol} & 1.3 \\
\hline
\end{array}
\][/tex]
The calculated density (0.934 g/cm³) is very close to the density of methyl acetate (0.93 g/cm³). Given the small disparity, a minor measurement error could account for the difference.
5. Conclusion:
Based on the density calculation, the unknown liquid is most likely methyl acetate.
