How much heat is required to vaporize 1.00 L of CCl4 at its normal boiling point?

The following data are given for CCl4.
Normal melting point, -23 degree celcius ,
normal boiling point, 77 degree celcius
density of liquid 1.59 g/mL; ;
enthalpy of fusion 3.28 kJ/mol
vapor pressure at 25 degree celcius ,
110 Torr

To calculate the heat required to vaporize 1.00 L of CCl4 at its normal boiling point, you need to consider the enthalpy of vaporization, which is the heat required to convert one mole of a substance from its liquid phase to its gas phase at its boiling point.

However, the enthalpy of vaporization is not directly provided in the given data. Instead, some other information is given. Let's break down the solution into steps:

Step 1: Calculate the number of moles of CCl4 in 1.00 L.
To do this, we need the density of the liquid. The density of CCl4 is given as 1.59 g/mL, which means 1.00 L of CCl4 has a mass of (1.59 g/mL) x (1000 mL) = 1590 g.

Now, we need to convert the mass into moles. The molar mass of CCl4 is:
(1 x 12.01 g/mol) + (4 x 35.45 g/mol) = 153.82 g/mol.

So, the number of moles is:
1590 g / 153.82 g/mol = 10.35 mol.

Step 2: Use the information about the enthalpy of fusion.
The enthalpy of fusion of CCl4 is given as 3.28 kJ/mol. This value represents the heat required to melt one mole of CCl4.

Even though we are looking for the enthalpy of vaporization, the enthalpy of fusion can provide some insight. We can assume that the heat of vaporization is roughly double the enthalpy of fusion because both processes involve breaking intermolecular forces.

Therefore, we can estimate the enthalpy of vaporization as:
2 x 3.28 kJ/mol = 6.56 kJ/mol.

Step 3: Calculate the heat required to vaporize 1.00 L of CCl4.
We know that 10.35 mol of CCl4 is present in 1.00 L. Therefore, the heat required to vaporize it would be:
10.35 mol x 6.56 kJ/mol = 67.78 kJ.

So, approximately 67.78 kJ of heat would be required to vaporize 1.00 L of CCl4 at its normal boiling point.