The heat of vaporization of benzene, C6H6, is 30.8kJ/mol at its boiling point of 80.1 degrees celsius. How much energy in the form of heat is required to vaporize 102g of benzene at its boiling point?

I know that the answer is 24.2kJ but i am not sure how to get this.

Beb

ano sagot?

To calculate the amount of energy required to vaporize a given mass of benzene, we can use the equation:

q = m * ΔHvap

Where:
- q is the amount of heat energy required (in joules)
- m is the mass of benzene (in grams)
- ΔHvap is the heat of vaporization of benzene (in joules per gram)

First, we need to convert the given mass of benzene from grams to moles. We can use the molar mass of benzene to do this:

Molar mass of C6H6 = (12.01 g/mol * 6) + (1.01 g/mol * 6) = 78.11 g/mol

Mass of benzene in moles = 102 g / 78.11 g/mol = 1.307 mol

Next, we can use the conversion factor of heat of vaporization to determine the amount of energy required to vaporize the given mass of benzene:

ΔHvap = 30.8 kJ/mol
ΔHvap = 30.8 kJ/1 mol

Now, we can calculate the amount of energy required:

q = m * ΔHvap
q = 1.307 mol * 30.8 kJ/mol
q = 40.1946 kJ

Finally, we need to convert the answer from kilojoules to kilojoules rounded to one decimal place:

q = 40.1946 kJ ≈ 40.2 kJ

So, the amount of energy required to vaporize 102g of benzene at its boiling point is approximately 40.2 kJ.

how many moles of benzene is in 102 g? Figure that out.

Now, energy=30.8kJ/mol * number of moles

which is not your "answer", which is wrong.