Questions: Calculating mass density metnyracetate: 0.93 mL/m pentane: 0.63 g/mL glycerol: 1.3 g/mL Next, the chemist measures the volume of the unknown liquid as 1645 cm^3 and the mass of the unknown liquid as 1.30 kg. Calculate the density of the liquid. Round your answer to 3 significant digits: g / mL Given the data above, is it possible to identify the liquid? - yes - no If it is possible to identify the liquid, do so: - ethanolamine - dimethyl sulfoxide - methyl acetate - pentane - Glycerol

Calculating mass density metnyracetate: 0.93 mL/m pentane: 0.63 g/mL glycerol: 1.3 g/mL Next, the chemist measures the volume of the unknown liquid as 1645 cm^3 and the mass of the unknown liquid as 1.30 kg. Calculate the density of the liquid. Round your answer to 3 significant digits: g / mL Given the data above, is it possible to identify the liquid? - yes - no If it is possible to identify the liquid, do so: - ethanolamine - dimethyl sulfoxide - methyl acetate - pentane - Glycerol
Transcript text: Calculating mass density \begin{tabular}{|c|c|} \hline metnyracetate & $0.93 \frac{\mathrm{~mL}}{\mathrm{~m}}$ \\ \hline pentane & $0.63 \frac{\mathrm{~g}}{\mathrm{~mL}}$ \\ \hline glycerol & $1.3 \frac{\mathrm{~g}}{\mathrm{~mL}}$ \\ \hline \end{tabular} Next, the chemist measures the volume of the unknown liquid as $1645 . \mathrm{cm}^{3}$ and the mass of the unknown liquid as 1.30 kg . \begin{tabular}{|c|c|} \hline Calculate the density of the liquid. Round your answer to 3 significant digits. & $\square \mathrm{g} / \mathrm{mL}$ \\ \hline Given the data above, is it possible to identify the liquid? & \begin{tabular}{l} O yes \\ 0 no \end{tabular} \\ \hline If it is possible to identify the liquid, do so. & ethanolamine dimethyl sulfoxide methyl acetate pentane Glycerol \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Calculate the density of the unknown liquid

To find the density of the unknown liquid, we use the formula: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] Given:

  • Mass = 1.30 kg = 1300 g (since 1 kg = 1000 g)
  • Volume = 1645 cm\(^3\) = 1645 mL (since 1 cm\(^3\) = 1 mL)

Substituting the values: \[ \text{Density} = \frac{1300 \, \text{g}}{1645 \, \text{mL}} \approx 0.7902 \, \text{g/mL} \]

Step 2: Determine if it is possible to identify the liquid

We compare the calculated density with the given densities of the substances:

  • Methyl acetate: \(0.93 \, \text{g/mL}\)
  • Pentane: \(0.63 \, \text{g/mL}\)
  • Glycerol: \(1.3 \, \text{g/mL}\)

The calculated density \(0.7902 \, \text{g/mL}\) does not match any of the given densities exactly.

Step 3: Conclusion on identifying the liquid

Since the calculated density does not match any of the given densities, it is not possible to identify the liquid based on the provided data.

Final Answer

\[ \boxed{\text{Density} = 0.790 \, \text{g/mL}} \] \[ \boxed{\text{No, it is not possible to identify the liquid}} \]

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