{\begin{tabular}{|c|c|} \hline liquid & density \ \hline glycerol & $1.3 g \cdot mL ^{-1}$ \ \hline dimethyl sulfoxide & $1.1 g \cdot mL ^{-1}$ \ \hline acetone & $0.79 g \cdot mL ^{-1}$ \ \hline methyl acetate & $0.93 g \cdot mL ^{-1}$ \ \hline ethanolamine & $1.0 g \cdot mL ^{-1}$ \ \hline \end{tabular}} Next, the chemist measures the volume of the unknown liquid as $1574 . cm ^3$ and the mass of the unknown liquid as {\begin{tabular}{|l|l|} \hline Calculate the density of the liquid. Round your answer to 3 significant digits. & $\square$ $g \cdot mL ^{-1}$ \ \hline Given the data above, is it possible to identify the liquid? & yes no \ \hline If it is possible to identify the liquid, do so. & glycerol dimethyl sulfoxide acetone methyl acetate ethanolamine \ \hline \end{tabular}}
Answer (2)
The calculated density of the unknown liquid is 1.0 g/mL, and based on this density, the liquid can be identified as ethanolamine. The density falls within the known values from the provided table. Thus, we can conclusively state the liquid's identity.
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Calculate the density using the formula: d e n s i t y = v o l u m e ma ss = 1574 m L 1574 g = 1.0 g / m L .
Round the calculated density to 3 significant digits, which remains 1.0 g / m L .
Compare the calculated density with the densities in the table.
Identify the liquid as ethanolamine, since its density matches the calculated density. The final answer is 1.0 g "." m L − 1 and the liquid is e t han o l amin e .
Explanation
Calculate the Density First, we need to calculate the density of the unknown liquid. We are given the volume as 1574 c m 3 . Since 1 c m 3 = 1 m L , the volume is also 1574 m L . We are also given the mass of the liquid as 1574 g .
Density Calculation To find the density, we use the formula: d e n s i t y = v o l u m e ma ss . Substituting the given values, we have: d e n s i t y = 1574 m L 1574 g = 1.0 g / m L .
Rounding the Density Now, we need to round the density to 3 significant digits. Since the calculated density is exactly 1.0 g / m L , it is already expressed with the required precision.
Identifying the Liquid Next, we compare the calculated density with the densities of the given liquids in the table to see if we can identify the unknown liquid. The table shows that ethanolamine has a density of 1.0 g / m L .
Final Answer Since the calculated density of the unknown liquid matches the density of ethanolamine, we can identify the liquid as ethanolamine.
Examples
Understanding density is crucial in many real-world applications. For instance, when designing ships, engineers need to consider the density of the materials used to ensure the ship floats. Similarly, in cooking, the order in which you add ingredients can depend on their densities, affecting how they layer or mix. In environmental science, density helps in understanding how pollutants spread in water or air. By calculating density, we can predict the behavior of substances in various scenarios, from everyday tasks to complex engineering projects.
