Benzene (C6H6) is an organic compound that boils at 80.1°C. If 100.0 grams of liquid benzene at 18.5°C is given 16.88 kJ of energy, what is the final temperature (in °C) of the benzene? The ΔHvaporization for benzene = 31.0 kJ/mol and the specific heat of liquid benzene = 2.18 J/gK.

To find the final temperature of benzene after it receives 16.88 kJ of energy, we need to consider two processes: heating the benzene to its boiling point and then vaporizing it.

First, let's calculate the heat required to raise the temperature of the benzene from 18.5°C to its boiling point of 80.1°C.

The specific heat formula is Q = mcΔT, where:
Q is the heat energy
m is the mass
c is the specific heat capacity
ΔT is the temperature change

Given:
Initial temperature, T1 = 18.5°C
Final temperature, T2 = 80.1°C
Mass, m = 100.0 grams
Specific heat capacity, c = 2.18 J/gK

Calculating the heat required to raise the temperature:
Q1 = mcΔT1
Q1 = (100.0 g) x (2.18 J/gK) x (80.1°C - 18.5°C)

Next, we need to calculate the heat required for vaporization using the enthalpy of vaporization (ΔHvaporization).

The formula for heat of vaporization is Q = nΔHvaporization, where:
Q is the heat energy
n is the number of moles of the substance
ΔHvaporization is the enthalpy of vaporization

We need to find the number of moles of benzene. The molar mass of benzene (C6H6) is 78.11 g/mol.

Number of moles (n) = mass (m) / molar mass (M)

n = 100.0 g / 78.11 g/mol

Finally, let's calculate the heat required for vaporization:
Q2 = n x ΔHvaporization
Q2 = (n) x (31.0 kJ/mol)

The total heat energy supplied to the benzene is the sum of Q1 and Q2:
Qtot = Q1 + Q2 = (Q1) + (Q2)

Now, let's calculate the total heat energy:
Qtot = (Q1) + (Q2) = [(100.0 g) x (2.18 J/gK) x (80.1°C - 18.5°C)] + [(100.0 g / 78.11 g/mol) x (31.0 kJ/mol)]

Once we have the total heat energy, we can solve for the final temperature T2 using the specific heat formula:
Qtot = mcΔT
T2 = [(Qtot) / (mc)] + T1

Plug in the values of Qtot, m, c, and T1, and calculate T2.