Question

Calculate the enthalpy of vaporization for a compound Z (mw = 86 g/mol) in kJ/mol if 448 kJ are required to vaporize 292 g of Z.

a. 131,000
b. 0.0178
c. 0.00758
d. 132
e. 0.000657

Answer 1

To find the enthalpy of vaporization, we divide the total energy required to vaporize the substance by the number of moles of the substance. The resultant value is the enthalpy of vaporization per mole, which in this case is 132 kJ/mol, making option d the correct answer.

Step-by-step explanation:

To calculate the enthalpy of vaporization for compound Z with a molecular weight (mw) of 86 g/mol, we need to use the energy required to vaporize the substance and the mass of the substance being vaporized. Given that 448 kJ of energy is required to vaporize 292 g of Z, we first convert the mass of Z to moles. The conversion is as follows:

Number of moles (n) = Mass (m) / Molecular weight (mw)
= 292 g / 86 g/mol
= 3.3953 mol

Now, we can calculate the enthalpy of vaporization per mole using the formula:

Enthalpy of vaporization (ΔHvap) = Total energy (Q) / Number of moles (n)
= 448 kJ / 3.3953 mol
We get the enthalpy of vaporization to be 131.96 kJ/mol, which rounds off to 132 kJ/mol when using significant figures.

Thus, the correct option is d. 132.