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A chemist working as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the cabinet is used to store bottles of acetone, methyl acetate, pentane, chloroform, and glycerol.

The chemist plans to try to identify the unknown liquid by measuring the density and comparing it to known densities. First, from his collection of Material Safety Data Sheets (MSDS), the chemist finds the following information:

\begin{tabular}{|c|c|}
\hline
liquid & density \\
\hline
acetone & [tex]$0.79 \, g \cdot cm^{-3}$[/tex] \\
\hline
methyl acetate & [tex]$0.93 \, g \cdot cm^{-3}$[/tex] \\
\hline
pentane & [tex]$0.63 \, g \cdot cm^{-3}$[/tex] \\
\hline
chloroform & [tex]$1.5 \, g \cdot cm^{-3}$[/tex] \\
\hline
glycerol & [tex]$1.3 \, g \cdot cm^{-3}$[/tex] \\
\hline
\end{tabular}

Next, the chemist measures the volume of the unknown liquid as [tex]$0.890 \, L$[/tex] and the mass of the unknown liquid as [tex]$831 \, g$[/tex].

1. Calculate the density of the liquid. Round your answer to 3 significant digits.

Density: [tex]$ \_\_\_\_\_\_ \, g \cdot cm^{-3}$[/tex]

2. Given the data above, is it possible to identify the liquid?

A. Yes
B. No

3. If it is possible to identify the liquid, do so:

A. Acetone
B. Methyl acetate
C. Pentane
D. Chloroform
E. Glycerol

Sagot :

GinnyAnswer
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.
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