The chemist measures the volume of the unknown liquid as 0.800L and the mass of the unknown liquid as 747g
Tetrahydrofuran 0.89g/cm^3
Glycerol 1.3g/cm^3
Dimethyl sulfoxide 1.1g/cm^3
Acetone 0.79 g/cm^3
Ethanolamine 1.0g/cm^3
Caluculate the density of liquid be sure your answer has the correct number of significant digits
Need answer in g/cm^3
Given data is it possible to identify the liquid
If it is possible to identify the liquid do so
To calculate the density of the unknown liquid, we need to use the formula: Density = Mass/Volume.
Given:
Volume of unknown liquid = 0.800 L
Mass of unknown liquid = 747 g
Density = 747g/0.800L = 933.75 g/L
To convert the density from g/L to g/cm^3, we need to divide by 1000 (since there are 1000 cm^3 in 1 L):
Density = 933.75 g/L / 1000 cm^3/L = 0.93375 g/cm^3
According to the given options, the closest density to 0.93375 g/cm^3 is for Ethanolamine, which has a density of 1.0 g/cm^3. Therefore, it is likely that the unknown liquid is Ethanolamine.
To calculate the density of the unknown liquid, we can use the formula:
Density = Mass / Volume
Given:
Volume (V) = 0.800 L
Mass (M) = 747 g
Density = 747 g / 0.800 L
Density ≈ 933.75 g/L
Now, let's convert the density to g/cm^3 by dividing it by 1000:
Density ≈ 933.75 g/L ÷ 1000 = 0.93375 g/cm^3
Now, let's compare the calculated density to the densities of the given liquids:
- Tetrahydrofuran: 0.89 g/cm^3
- Glycerol: 1.3 g/cm^3
- Dimethyl sulfoxide: 1.1 g/cm^3
- Acetone: 0.79 g/cm^3
- Ethanolamine: 1.0 g/cm^3
Since the calculated density (0.93375 g/cm^3) is closest to the density of Dimethyl sulfoxide (1.1 g/cm^3), it is likely that the unknown liquid is Dimethyl sulfoxide.