How much heat is required to vaporize 3.50L of at its normal boiling point?

The following data are given for Carbon Tetrachloride. Normal melting point, -23 Degrees Celsius, normal boiling point, 77 Degrees Celsius; density of liquid, 1.59g/mL; Delta H (FUSION): 3.28 kJ mol^-1; vapor pressure at 25 Degrees Celsius, 110 Torr.

**I don't know how to find the heat of vaporization, though when the boiling point is 77 degrees celsius, I found that the heat of vaporization if 192J/g, allowing the FINAL answer to equal to 1068.48kJ, because I was told to be wrong, can someone please correct me?

Thank you.

Have you tried using the Clausius-Clapeyron equation to find delta H vap? You have P1 and P2, and T1 and T2. Or is that how you found 192 J/g?

And I don't see a starting point for temperature. If you are to determine total heat required from some T to vaporization, you must have both delta H vap and a starting T.

I found 32.54 kJ/mol for delta H vap (on the Internet).

thank you!!!

Secondly, how is it that you solve this type of question, I have no clue in finding a way to start the problem.

thanks

Do you have a temperature from which to start? Said another way, are you to determine the total heat from some temperature up to and including the vaporization OR are you to determine only the heat from the vaporization?

I am to find how much heat is required to vaporize 3.50L of at its normal boiling point?

such that the normal boiling point: 77 Degrees Celsius

and the following data are provided,
Normal melting point, -23 Degrees Celsius
density of liquid, 1.59g/mL
Delta H (FUSION): 3.28 kJ mol^-1
vapor pressure at 25 Degrees Celsius, 110 Torr

Thanks.

If you need only the heat required to vaporize it, then most of the data provided is not useful (heat fusion, etc).

First, convert 3.50 L to grams using the density listed.
I looked up the heat of vap on the web and found 32.54 kJ/mol. Change that to J/gram
32.54 kJ/mol x (1 mol/molar mass CCl4) x (1000 J/kJ) = heat vap in J/g.
Now
mass CCl4 x heat vap in J/g = ??
Check my thinking. Check my work.

Yes!!!

Thank you very much for guiding through this question!

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To find the amount of heat required to vaporize 3.50L of carbon tetrachloride at its normal boiling point, you need to use the equation:

q = n * ΔHvap

where:
q is the heat required (in joules)
n is the number of moles of carbon tetrachloride
ΔHvap is the molar heat of vaporization

First, you need to calculate the number of moles of carbon tetrachloride in 3.50L. To do this, you can use the ideal gas law:

PV = nRT

where:
P is the pressure
V is the volume
n is the number of moles
R is the gas constant
T is the temperature

In this case, the carbon tetrachloride is in liquid form, so we need to use the vapor pressure at 77 degrees Celsius (its boiling point). Convert this temperature to Kelvin by adding 273.15:

77 + 273.15 = 350.15K

The boiling point is given as the temperature at which the vapor pressure equals atmospheric pressure (1 atm or 760 Torr). Convert this to Torr:

1 atm = 760 Torr

Now we can use the ideal gas law to calculate the number of moles:

(760 Torr) * (3.50L) = n * (0.0821 L * atm/(mol * K)) * (350.15K)

Solving for n, we get:

n = (760 Torr * 3.50L) / (0.0821 L * atm/(mol * K) * 350.15K)

n ≈ 88.6 mol

Now that we have the number of moles, we can calculate the heat required using the molar heat of vaporization.

ΔHvap = 192 J/g = 192 kJ/mol (since 1 g = 1 mol for carbon tetrachloride)

q = (88.6 mol) * (192 kJ/mol)

q ≈ 17,011.2 kJ

Therefore, the amount of heat required to vaporize 3.50L of carbon tetrachloride at its normal boiling point is approximately 17,011.2 kJ, not 1068.48 kJ as you mentioned.