Answer :
Sure, let's go through the steps to solve this problem.
### Step 1: Calculating the Density
We are given:
- Mass of the unknown liquid = 472 grams
- Volume of the unknown liquid = 0.597 liters
First, we need to convert the volume from liters to cubic centimeters because the given densities are in grams per cubic centimeter (g/cm³).
1 liter = 1000 cubic centimeters (cm³).
So,
[tex]\[ \text{Volume in cm}^3 = 0.597 \, \text{L} \times 1000 \, \text{cm}^3/\text{L} = 597 \, \text{cm}^3 \][/tex]
Next, we calculate the density:
[tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]
[tex]\[ \text{Density} = \frac{472 \, \text{g}}{597 \, \text{cm}^3} \][/tex]
[tex]\[ \text{Density} \approx 0.7906197654941374 \, \text{g/cm}^3 \][/tex]
Now, we round this to three significant digits:
[tex]\[ \text{Density} \approx 0.791 \, \text{g/cm}^3 \][/tex]
### Step 2: Comparing with Known Densities
The known densities from the MSDS information are:
- Acetone: [tex]\( 0.79 \, \text{g/cm}^3 \)[/tex]
- Dimethyl sulfoxide: [tex]\( 1.1 \, \text{g/cm}^3 \)[/tex]
- Carbon tetrachloride: [tex]\( 1.6 \, \text{g/cm}^3 \)[/tex]
- Glycerol: [tex]\( 1.3 \, \text{g/cm}^3 \)[/tex]
- Methyl acetate: [tex]\( 0.93 \, \text{g/cm}^3 \)[/tex]
We compare the calculated density (0.791 g/cm³) to the given densities:
- 0.791 is closest to 0.79 (the density of acetone).
### Step 3: Answer the Questions
1. Calculate the density of the liquid. Round your answer to 3 significant digits.
[tex]\[ \boxed{0.791 \, \text{g/cm}^3} \][/tex]
2. Given the data above, is it possible to identify the liquid?
[tex]\[ \boxed{\text{No}} \][/tex]
3. If it is possible to identify the liquid, do so.
In this case, even though the calculated density is very close to the density of acetone, it does not match any of the densities exactly, therefore it is not definitively possible to identify the liquid based on the given densities.
Therefore, the results suggest that:
- It is not possible to definitively identify the liquid based on the data provided. So, we leave this part unmarked (None).
### Step 1: Calculating the Density
We are given:
- Mass of the unknown liquid = 472 grams
- Volume of the unknown liquid = 0.597 liters
First, we need to convert the volume from liters to cubic centimeters because the given densities are in grams per cubic centimeter (g/cm³).
1 liter = 1000 cubic centimeters (cm³).
So,
[tex]\[ \text{Volume in cm}^3 = 0.597 \, \text{L} \times 1000 \, \text{cm}^3/\text{L} = 597 \, \text{cm}^3 \][/tex]
Next, we calculate the density:
[tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]
[tex]\[ \text{Density} = \frac{472 \, \text{g}}{597 \, \text{cm}^3} \][/tex]
[tex]\[ \text{Density} \approx 0.7906197654941374 \, \text{g/cm}^3 \][/tex]
Now, we round this to three significant digits:
[tex]\[ \text{Density} \approx 0.791 \, \text{g/cm}^3 \][/tex]
### Step 2: Comparing with Known Densities
The known densities from the MSDS information are:
- Acetone: [tex]\( 0.79 \, \text{g/cm}^3 \)[/tex]
- Dimethyl sulfoxide: [tex]\( 1.1 \, \text{g/cm}^3 \)[/tex]
- Carbon tetrachloride: [tex]\( 1.6 \, \text{g/cm}^3 \)[/tex]
- Glycerol: [tex]\( 1.3 \, \text{g/cm}^3 \)[/tex]
- Methyl acetate: [tex]\( 0.93 \, \text{g/cm}^3 \)[/tex]
We compare the calculated density (0.791 g/cm³) to the given densities:
- 0.791 is closest to 0.79 (the density of acetone).
### Step 3: Answer the Questions
1. Calculate the density of the liquid. Round your answer to 3 significant digits.
[tex]\[ \boxed{0.791 \, \text{g/cm}^3} \][/tex]
2. Given the data above, is it possible to identify the liquid?
[tex]\[ \boxed{\text{No}} \][/tex]
3. If it is possible to identify the liquid, do so.
In this case, even though the calculated density is very close to the density of acetone, it does not match any of the densities exactly, therefore it is not definitively possible to identify the liquid based on the given densities.
Therefore, the results suggest that:
- It is not possible to definitively identify the liquid based on the data provided. So, we leave this part unmarked (None).