How much energy is required to vaporize 75.5 mL of ethanol (C2H5OH, d=0.789 g/cm^3) at its boiling point, if its heat of vaporization is 40.5 kJ/mol?
To calculate the energy required to vaporize ethanol, we can follow these steps:
Step 1: Calculate the mass of ethanol.
Given the volume of ethanol is 75.5 mL, and the density of ethanol is 0.789 g/cm^3, we can find the mass using the formula:
Mass = Volume × Density.
Mass = 75.5 mL × 0.789 g/cm^3
Step 2: Convert the mass from grams to moles.
To do this, we need the molar mass of ethanol (C2H5OH), which is calculated by summing the atomic masses of all the elements in the molecule.
The molar mass of carbon (C) = 12.01 g/mol.
The molar mass of hydrogen (H) = 1.008 g/mol.
The molar mass of oxygen (O) = 16.00 g/mol.
Molar mass of C2H5OH = (2 × 12.01 g/mol) + (6 × 1.008 g/mol) + (1 × 16.00 g/mol)
Step 3: Calculate the number of moles.
Divide the mass of ethanol (from Step 1) by the molar mass of ethanol (from Step 2).
Number of moles = Mass / Molar mass
Step 4: Calculate the energy required.
The heat of vaporization of ethanol is given as 40.5 kJ/mol.
Energy required = Heat of vaporization × Number of moles
Now let's calculate:
Step 1: Mass = 75.5 mL × 0.789 g/cm^3
Step 2: Molar mass of C2H5OH = (2 × 12.01 g/mol) + (6 × 1.008 g/mol) + (1 × 16.00 g/mol)
Step 3: Number of moles = Mass / Molar mass
Step 4: Energy required = Heat of vaporization × Number of moles
To find the energy required to vaporize a given quantity of ethanol, we need to use the following formula:
Energy = (mass of ethanol) x (heat of vaporization)
First, let's calculate the mass of ethanol using the given volume and density.
The formula to find the mass of an object is:
mass = volume x density
Given data:
Volume of ethanol (V) = 75.5 mL
Density of ethanol (d) = 0.789 g/cm^3
To use the same units, we need to convert mL to cm³:
1 mL = 1 cm³
So, the volume of ethanol in cm³ is 75.5 cm³.
Now, we can calculate the mass using the formula:
mass = volume x density
mass = 75.5 cm³ x 0.789 g/cm³
mass = 59.5945 g (rounded to 4 decimal places)
Now that we have the mass of ethanol, we can calculate the energy required to vaporize it using the given heat of vaporization.
Given data:
Heat of vaporization of ethanol (ΔHvap) = 40.5 kJ/mol
Note that the heat of vaporization is given per mole of ethanol, so we need to convert the mass of ethanol to moles.
First, let's find the molar mass of ethanol (C₂H₅OH):
C = 12.01 g/mol (atomic mass of carbon)
H = 1.01 g/mol (atomic mass of hydrogen)
O = 16.00 g/mol (atomic mass of oxygen)
Molar mass of ethanol (C₂H₅OH) = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.07 g/mol
Now, we can convert the mass of ethanol to moles:
moles = mass / molar mass
moles = 59.5945 g / 46.07 g/mol
moles ≈ 1.2941 mol (rounded to 4 decimal places)
Finally, we can calculate the energy required to vaporize the ethanol using the formula:
Energy = (mass of ethanol) x (heat of vaporization)
Energy = 1.2941 mol x 40.5 kJ / mol
Energy ≈ 52.42 kJ (rounded to 2 decimal places)
Therefore, the energy required to vaporize 75.5 mL of ethanol at its boiling point is approximately 52.42 kJ.