How much heat is required to vaporize 343g of liquid ethanol at its boiling point?

ΔHvap+38.6 Kj\mol

To calculate the amount of heat required to vaporize a given mass of a substance, you can use the equation:

q = m × ΔHvap

Where:
q = amount of heat (in joules)
m = mass of the substance (in grams)
ΔHvap = molar heat of vaporization (in joules per mole)

First, let's convert the mass of ethanol from grams to moles:


molar mass of ethanol (C2H5OH) = (2 × atomic mass of carbon) + (6 × atomic mass of hydrogen) + atomic mass of oxygen
≈ (2 × 12.01 g/mol) + (6 × 1.01 g/mol) + 16.00 g/mol
≈ 46.07 g/mol

Number of moles of ethanol = mass of ethanol / molar mass of ethanol
= 343 g / 46.07 g/mol
≈ 7.453 mol

Next, we can calculate the amount of heat required using the equation:

q = m × ΔHvap

where:
m = 7.453 mol (converted mass of ethanol to moles)
ΔHvap = 38.6 kJ/mol (given molar heat of vaporization)

Now we will convert the units of ΔHvap from kilojoules to joules:

ΔHvap = 38.6 kJ/mol × 1000 J/kJ
≈ 38600 J/mol

Finally, we can calculate the amount of heat required:

q = 7.453 mol × 38600 J/mol
≈ 287,823.8 J

Therefore, approximately 287,823.8 joules of heat is required to vaporize 343 grams of liquid ethanol at its boiling point.

To calculate the amount of heat required to vaporize a given quantity of a substance at its boiling point, we can use the equation:

Q = m * ΔHvap

Where Q is the heat required, m is the mass of the substance, and ΔHvap is the molar heat of vaporization.

First, we need to convert the mass of ethanol from grams to moles. The molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol, so:

Number of moles (n) = mass / molar mass
= 343g / 46.07 g/mol

Next, we can calculate the heat required using the equation mentioned above:

Q = n * ΔHvap

But before we proceed, we need to convert the molar heat of vaporization from kJ/mol to J/mol, as the mass of ethanol we have is in grams:

ΔHvap = 38.6 kJ/mol * 1000 J/1 kJ
= 38,600 J/mol

Now we can substitute the values into the equation:

Q = (343g / 46.07 g/mol) * (38,600 J/mol)

By performing this calculation, we can determine the amount of heat (Q) required to vaporize 343g of liquid ethanol at its boiling point.

q = mass x deltaHvap. Watch the units, you must change either 343 g to moles or change delta H vap to kJ/g

You are STUPID!