A 15.0g sample of liquid ethanol, C2H5OH, absorbs 5.13x10^3 J of heat at its normal boiling point, 78.0C. The molar enthalpy of vaporization of ethanol is 39.3kJ/mol. (a) What volume of ethanol vapor is produced? The voume is measured at 78.0C and 1.00atm pressure. (b) What mass of ethanol remians in the liquid state?
To find the answers to these questions, we need to use the information given and apply the concept of molar enthalpy of vaporization.
(a) To find the volume of ethanol vapor produced, we first need to convert the mass of ethanol into moles. We can use the molar mass of ethanol to do this.
The molar mass of ethanol (C2H5OH) can be calculated as follows:
(2 × Atomic mass of carbon) + (6 × Atomic mass of hydrogen) + Atomic mass of oxygen
The atomic masses are:
Carbon (C): 12.01 g/mol
Hydrogen (H): 1.01 g/mol
Oxygen (O): 16.00 g/mol
The molar mass of ethanol would be:
(2 × 12.01 g/mol) + (6 × 1.01 g/mol) + 16.00 g/mol = 46.08 g/mol
Using this molar mass, we can calculate the number of moles in the given 15.0 g sample of ethanol:
Number of moles = Mass of ethanol / Molar mass of ethanol
Number of moles = 15.0 g / 46.08 g/mol
Now that we have the number of moles of ethanol, we can use the molar enthalpy of vaporization to find the energy required to vaporize the sample.
The molar enthalpy of vaporization given is 39.3 kJ/mol. However, we need to convert this value to J/mol to match the energy absorbed.
1 kJ = 1000 J
Therefore, the molar enthalpy of vaporization in J/mol is:
39.3 kJ/mol × 1000 J/1 kJ = 39300 J/mol.
Using this value, we can calculate the total energy required to vaporize the sample:
Total energy = Number of moles of ethanol × Molar enthalpy of vaporization
Total energy = (15.0 g / 46.08 g/mol) × 39300 J/mol
Now that we have the total energy required, we can use the given energy absorbed at the boiling point (5.13 × 10^3 J) to find the number of moles of ethanol vaporized.
Number of moles of ethanol vaporized = Energy absorbed / Total energy required
Once we have the number of moles of ethanol vaporized, we can use the ideal gas law to find the volume of ethanol vapor produced.
The ideal gas law equation is:
PV = nRT
Since we are given the pressure (1.00 atm), temperature (78.0°C), and the number of moles of ethanol vapor (from the previous step), we can calculate the volume using the ideal gas law.
(b) To find the mass of ethanol remaining in the liquid state, we need to subtract the mass of the ethanol vaporized from the initial mass of the liquid ethanol (15.0 g).
Mass of ethanol remaining = Initial mass of ethanol - Mass of ethanol vaporized
By following these steps, you should be able to find the answers to both questions.