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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 methyl acetate, carbon tetrachloride, diethylamine, glycerol, and chloroform.

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}{|r|c|}
\hline
Liquid & Density \\
\hline
Methyl acetate & [tex]$0.93 \frac{ g }{ mL }$[/tex] \\
\hline
Carbon tetrachloride & [tex]$1.6 \frac{ g }{ mL }$[/tex] \\
\hline
Diethylamine & [tex]$0.71 \frac{ g }{ mL }$[/tex] \\
\hline
Glycerol & [tex]$1.3 \frac{ g }{ mL }$[/tex] \\
\hline
Chloroform & [tex]$1.5 \frac{ g }{ mL }$[/tex] \\
\hline
\end{tabular}

Next, the chemist measures the volume of the unknown liquid as [tex]$1115 \, \text{cm}^3$[/tex] and the mass of the unknown liquid as 1.65 kg.

\begin{tabular}{|l|l|}
\hline
Calculate the density of the liquid. Round your answer to 3 significant digits. & \underline{\hspace{2cm}} [tex]$\frac{ g }{ mL }$[/tex] \\
\hline
Given the data above, is it possible to identify the liquid? & \underline{\hspace{1cm}} Yes \underline{\hspace{1cm}} No \\
\hline
If it is possible to identify the liquid, do so. & \underline{\hspace{2cm}} Methyl acetate \\
& \underline{\hspace{2cm}} Carbon tetrachloride \\
& \underline{\hspace{2cm}} Diethylamine \\
& \underline{\hspace{2cm}} Glycerol \\
& \underline{\hspace{2cm}} Chloroform \\
\hline
\end{tabular}



Answer :

GinnyAnswer
To solve this problem, we need to calculate the density of the unknown liquid and then compare it to the known densities of the substances listed.

Step-by-Step Solution:

1. Convert the mass from kilograms to grams:
- Given mass: [tex]\( 1.65 \text{ kg} \)[/tex]
- Since [tex]\( 1 \text{ kg} = 1000 \text{ grams} \)[/tex]
- Therefore, mass in grams: [tex]\( 1.65 \times 1000 = 1650 \text{ grams} \)[/tex]

2. Calculate the density:
- Density is defined as mass divided by volume.
- Given volume: [tex]\( 1115 \text{ cm}^3 \)[/tex]
- Density of the liquid:
[tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{1650 \text{ grams}}{1115 \text{ cm}^3} \][/tex]
- Perform the division:
[tex]\[ \text{Density} \approx 1.480 \frac{\text{grams}}{\text{cm}^3} \][/tex]
- Round the result to 3 significant digits:
[tex]\[ 1.48 \frac{\text{grams}}{\text{cm}^3} \][/tex]

3. Compare the calculated density with the given densities:
- Methyl acetate: [tex]\( 0.93 \frac{\text{g}}{\text{mL}} \)[/tex]
- Carbon tetrachloride: [tex]\( 1.6 \frac{\text{g}}{\text{mL}} \)[/tex]
- Diethylamine: [tex]\( 0.71 \frac{\text{g}}{\text{mL}} \)[/tex]
- Glycerol: [tex]\( 1.3 \frac{\text{g}}{\text{mL}} \)[/tex]
- Chloroform: [tex]\( 1.5 \frac{\text{g}}{\text{mL}} \)[/tex]

4. Identify the closest match:
- The calculated density is [tex]\( 1.48 \frac{\text{g}}{\text{mL}} \)[/tex].
- Compared to the given densities, it is closest to chloroform which has a density of [tex]\( 1.5 \frac{\text{g}}{\text{mL}} \)[/tex].

Conclusion:
- The density of the unknown liquid is [tex]\( 1.48 \frac{\text{g}}{\text{mL}} \)[/tex].
- Based on the densities provided in the MSDS, the unknown liquid is most likely chloroform.