When you want to calculate the energy absorbed when a mass of liquid boils, you need to consider the enthalpy change associated with the phase transition from liquid to gas. This enthalpy change is known as the heat of vaporization (ΔHvap), which represents the energy required to turn one mole of a liquid into gas at its boiling point under constant pressure.
Stoichiometry involves using the relationships between the quantities of reactants and products in a chemical reaction to calculate various quantities. For this calculation, the relevant relationship involves converting the mass of the liquid to moles, and then using the heat of vaporization to find the energy absorbed.
Here’s a step-by-step approach:
1. Convert grams of liquid to moles:
- The number of moles ([tex]\( n \)[/tex]) can be found using the molar mass ([tex]\( M \)[/tex]) of the liquid. The formula is:
[tex]\[
n = \frac{\text{grams of liquid}}{\text{molar mass of the liquid}}
\][/tex]
2. Calculate the energy absorbed using the heat of vaporization:
- Once you have the number of moles, the energy absorbed ([tex]\( Q \)[/tex]) can be calculated by multiplying the number of moles by the heat of vaporization ([tex]\( \Delta H_{\text{vap}} \)[/tex]).
[tex]\[
Q = n \times \Delta H_{\text{vap}}
\][/tex]
So, the correct formula to calculate the energy absorbed when a liquid boils is:
[tex]\[
\text{Grams liquid} \times \frac{\text{mol}}{\text{g}} \times \Delta H_{\text{vap}}
\][/tex]
Given this explanation and knowing that the heat of vaporization ([tex]\( \Delta H_{\text{vap}} \)[/tex]) is the relevant enthalpy for boiling, the correct option is:
C. Grams liquid × mol / g × ΔHvap
