(a) A sample of tetrachloroethylene, a liquid used in dry cleaning that is being phased out because of its potential to cause cancer, has a mass of 40.55 \(\mathrm{g}\) and a volume of 25.0 \(\mathrm{mL}\) at \(25^{\circ} \mathrm{C}\) . What is its density at this temperature? Will tetrachloroethylene float on water? (Materials that are less dense than water will float.) (b) Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a usupercritical fluid" that is a much safer dry-cleaning agent than tetrachlorosthvlene, At a certain pressure, the density of super critical \(\mathrm{CO}_{2}\) is 0.469 \(\mathrm{g} / \mathrm{cm}^{3} .\) What is the mass of a 25.0 \(\mathrm{-mL}\) sample of supercritical \(\mathrm{CO}_{2}\) at this pressure?

Step by step solution

01

Calculate the density of tetrachloroethylene

We are given the mass and volume of tetrachloroethylene. We can use the formula for density to find its value: Density = \(\frac{Mass}{Volume}\) We can plug in the given values to obtain the density: Density = \(\frac{40.55 \,\text{g}}{25.0 \,\text{mL}}\) Now, we calculate the density: Density = \(1.622 \, \frac{\text{g}}{\text{mL}}\)

02

Compare density with water

To determine if tetrachloroethylene will float on water, we must compare its density with the density of water. The density of water is approximately \(1.0 \, \frac{\text{g}}{\text{mL}}\) at \(25^{\circ} \mathrm{C}\). Since the density of tetrachloroethylene (\(1.622 \, \frac{\text{g}}{\text{mL}}\)) is greater than the density of water (\(1.0 \, \frac{\text{g}}{\text{mL}}\)), it will not float on water.

03

Calculate the mass of supercritical CO2

We are given the density and volume of supercritical CO2, and we need to calculate its mass. We can rearrange the density formula to get the desired value: Mass = Density × Volume Plugging in the given values, we can calculate the mass of supercritical CO2: Mass = \(0.469 \, \frac{\text{g}}{\text{cm}^3} \times 25.0 \,\text{mL}\) (Note that \(1 \, \text{mL} = 1 \, \text{cm}^3\).) Now, we calculate the mass: Mass = \(11.725 \, \text{g}\) Therefore, a 25.0 mL sample of supercritical CO2 at the given pressure has a mass of 11.725 g.

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