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.

To find the heat required to vaporize 3.50L of carbon tetrachloride (CCl4) at its normal boiling point, we need to consider several steps and equations involved.

First, let's calculate the number of moles of CCl4 in 3.50L using the ideal gas law:

PV = nRT

Where:
P = pressure (in atmosphere)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K for this problem)
T = temperature (in Kelvin)

Since the temperature is at the normal boiling point, which is 77 degrees Celsius (350K), and we know the vapor pressure at 25 degrees Celsius (298K), we can use the equation:

P1/T1 = P2/T2

Plugging in the values we have:
110 Torr / 298K = P2 / 350K

Solving for P2:
P2 = (110 Torr / 298K) * 350K
P2 ≈ 128.52 Torr

Now that we know the partial pressure of CCl4 vapor at its boiling point, we can use the ideal gas law to convert the vapor pressure to moles:

PV = nRT

Where:
P = pressure (in atmosphere)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K for this problem)
T = temperature (in Kelvin)

Plugging in the values we have:
(128.52 Torr / 760 Torr/atm) * 3.50L = n * 0.0821 L·atm/mol·K * 350K

Solving for n:
n = [(128.52 Torr / 760 Torr/atm) * 3.50L] / [0.0821 L·atm/mol·K * 350 K]
n ≈ 0.0861 moles

Now that we have the number of moles of CCl4, we can calculate the heat of vaporization (delta Hvap), which is the amount of heat required to vaporize one mole of CCl4. From the given data, the heat of fusion (delta Hfusion) is given as 3.28 kJ/mol.

However, you mentioned that the heat of vaporization at the boiling point is 192 J/g. To use this value, we need to convert moles to grams using the molecular weight of CCl4, which is 153.82 g/mol:

Mass = moles * molecular weight
Mass ≈ 0.0861 mol * 153.82 g/mol
Mass ≈ 13.26 g

Now we can calculate the heat of vaporization using the specific heat formula:

Heat = mass * delta Hvap

Heat = 13.26 g * 192 J/g
Heat ≈ 2549.92 J

Finally, we can convert the heat from joules to kilojoules. There are 1000 joules in 1 kilojoule:

Heat = 2549.92 J / 1000
Heat ≈ 2.55 kJ

Therefore, the correct answer for the heat required to vaporize 3.50L of CCl4 at its normal boiling point is approximately 2.55 kJ.