To calculate the density of the liquid and determine if it matches any of the known substances given, follow these steps:
### Step-by-Step Solution:
1. Record the Given Data:
- Volume of the liquid, [tex]\( V \)[/tex]: [tex]\( 1424 \, \text{cm}^3 \)[/tex]
- Mass of the liquid, [tex]\( m \)[/tex]: [tex]\( 1.33 \, \text{kg} \)[/tex]
2. Convert Mass from Kilograms to Grams:
- Note that [tex]\( 1 \, \text{kg} = 1000 \, \text{g} \)[/tex]
- Therefore, [tex]\( 1.33 \, \text{kg} = 1.33 \times 1000 \, \text{g} = 1330.0 \, \text{g} \)[/tex]
3. Calculate the Density:
- Density ([tex]\( \rho \)[/tex]) is defined as mass divided by volume.
- Formula: [tex]\( \rho = \frac{m}{V} \)[/tex]
- Substituting the known values:
[tex]\[
\rho = \frac{1330.0 \, \text{g}}{1424 \, \text{cm}^3}
\][/tex]
4. Perform the Division to Find the Density:
[tex]\[
\rho = \frac{1330.0}{1424} \, \text{g/cm}^3 \approx 0.9339887640449438 \, \text{g/cm}^3
\][/tex]
5. Round the Density to 3 Significant Digits:
- The density calculated above needs to be rounded to 3 significant digits.
- Consequently, density [tex]\( \rho \approx 0.934 \, \text{g/cm}^3 \)[/tex]
### Summary of Results:
- Volume: [tex]\( 1424 \, \text{cm}^3 \)[/tex]
- Mass: [tex]\( 1.33 \, \text{kg} \)[/tex]
- Converted Mass: [tex]\( 1330.0 \, \text{g} \)[/tex]
- Density (Exact): [tex]\( 0.9339887640449438 \, \text{g/cm}^3 \)[/tex]
- Density (Rounded to 3 Significant Digits): [tex]\( 0.934 \, \text{g/cm}^3 \)[/tex]
### Identification of the Liquid:
Given the density calculation, we can compare [tex]\( 0.934 \, \text{g/cm}^3 \)[/tex] to the typical densities of the provided substances. However, identifying the liquid solely based on density requires comparing it to a density table, which rules out the identification. Based on general knowledge:
- Common Densities:
- Acetone: [tex]\( \approx 0.79 \, \text{g/cm}^3 \)[/tex]
- Pentane: [tex]\( \approx 0.626 \, \text{g/cm}^3 \)[/tex]
- Chloroform: [tex]\( \approx 1.48 \, \text{g/cm}^3 \)[/tex]
- Glycerol: [tex]\( \approx 1.26 \, \text{g/cm}^3 \)[/tex]
- Methyl Acetate: [tex]\( \approx 0.932 \, \text{g/cm}^3 \)[/tex]
Given the value [tex]\( 0.934 \, \text{g/cm}^3 \)[/tex] closely matches the density of Methyl Acetate, but not exactly the other substances listed. However, for an absolute identification, more data or additional properties should be compared.
Therefore, based on just the density, we only have an indicative match rather than a precise identification.
