The Correct Answer and Explanation is:
Dichloromethane (CH2Cl2) has a density of 1.33 g/mL. What volume of vapor would be generated by the vaporization of 2.00 ml of dichloromethane at 100°C?
The Correct Answer and Explanation is:
To calculate the volume of vapor generated by the vaporization of 2.00 mL of dichloromethane (CH₂Cl₂) at 100°C, we follow these steps:
Step-by-step Calculation:
- Determine the mass of liquid dichloromethane:
Mass=Density×Volume=1.33 g/mL×2.00 mL=2.66 g\text{Mass} = \text{Density} \times \text{Volume} = 1.33 \, \text{g/mL} \times 2.00 \, \text{mL} = 2.66 \, \text{g}
- Convert mass to moles:
The molar mass of CH₂Cl₂ is: (12.01 C)+2(1.008 H)+2(35.45 Cl)=84.93 g/mol(12.01 \, \text{C}) + 2(1.008 \, \text{H}) + 2(35.45 \, \text{Cl}) = 84.93 \, \text{g/mol} Moles=2.66 g84.93 g/mol≈0.0313 mol\text{Moles} = \frac{2.66 \, \text{g}}{84.93 \, \text{g/mol}} \approx 0.0313 \, \text{mol}
- Use the Ideal Gas Law to find the vapor volume:
PV=nRT⇒V=nRTPPV = nRT \Rightarrow V = \frac{nRT}{P}
Where:
- n=0.0313 moln = 0.0313 \, \text{mol}
- R=0.0821 L\cdotpatm/mol\cdotpKR = 0.0821 \, \text{L·atm/mol·K}
- T=100°C=373.15 KT = 100°C = 373.15 \, \text{K}
- P=1.00 atmP = 1.00 \, \text{atm}
V=(0.0313)(0.0821)(373.15)1.00≈0.957 LV = \frac{(0.0313)(0.0821)(373.15)}{1.00} \approx 0.957 \, \text{L}
Answer:
Approximately 0.96 liters (or 960 mL) of vapor is produced.
Explanation
Dichloromethane (CH₂Cl₂), also known as methylene chloride, is a volatile organic compound that readily vaporizes under heat. When liquid CH₂Cl₂ is heated to 100°C, it transitions into the gas phase. The question involves determining the volume of vapor generated from 2.00 mL of the liquid at this elevated temperature and standard atmospheric pressure.
We begin by converting the volume of liquid to mass using its density (1.33 g/mL). This gives 2.66 grams of CH₂Cl₂. Next, we convert the mass to moles using its molar mass (approximately 84.93 g/mol), yielding about 0.0313 moles.
To determine the volume of gas produced, we apply the Ideal Gas Law (PV=nRTPV = nRT). This equation relates the pressure, volume, temperature, and number of moles of an ideal gas. Although real gases deviate slightly from ideal behavior, the approximation is sufficiently accurate under moderate conditions like 100°C and 1 atm.
We input the known values into the rearranged Ideal Gas Law V=nRTPV = \frac{nRT}{P}, with R being the gas constant (0.0821 L·atm/mol·K) and the temperature converted to Kelvin (373.15 K). Solving this gives us a vapor volume of approximately 0.96 liters.
This result illustrates how even small amounts of a volatile liquid can produce significantly larger volumes of gas upon vaporization, which is critical in chemical processes, lab safety, and environmental considerations involving volatile organic compounds.
