To determine the mass of liquid that can be boiled using 1 kJ of energy given the enthalpy of vaporization (ΔHvap), follow these detailed steps:
1. Understand the relationship between energy and vaporization:
- The enthalpy of vaporization (ΔHvap) is the amount of energy required to convert one mole of a liquid into its gaseous state at constant temperature and pressure.
- The unit of ΔHvap is typically given in kJ/mol, which means how many kilojoules are needed to vaporize one mole of the liquid.
2. Set up the formula:
- To find out how much mass of the liquid can be vaporized, you start with the amount of energy (1 kJ) and use the enthalpy of vaporization to determine the moles of liquid that can be vaporized.
3. Convert energy to moles:
- The formula to convert the energy to moles of liquid vaporized is given by:
[tex]\[
\text{moles} = \frac{\text{energy}}{\Delta H_{\text{vap}}}
\][/tex]
- For 1 kJ of energy, it simplifies to:
[tex]\[
\text{moles} = \frac{1 \text{ kJ}}{\Delta H_{\text{vap}}}
\][/tex]
4. Convert moles to mass:
- Once you have the number of moles vaporized, convert moles to mass using the molar mass of the liquid (molar mass is in g/mol).
- The formula to find the mass is:
[tex]\[
\text{mass} = \text{moles} \times \text{molar mass}
\][/tex]
- Since [tex]\(\text{moles} = \frac{1 \text{ kJ}}{\Delta H_{\text{vap}}} \)[/tex], substitute this into the mass formula:
[tex]\[
\text{mass} = \left( \frac{1 \text{ kJ}}{\Delta H_{\text{vap}}} \right) \times \text{molar mass}
\][/tex]
- Rearrange the terms to fit a multiplication format:
[tex]\[
\text{mass} = 1 \text{ kJ} \times \left( \frac{1}{\Delta Hvap} \right) \times \text{molar mass}
\][/tex]
The correct method to find the mass of liquid vaporized by 1 kJ of energy is:
[tex]\[
1 \text{ kJ} \times \left( \frac{1}{\Delta H_{\text{vap}}} \right) \times \text{g/mol} \text{ liquid}
\][/tex]
This matches option A.
Therefore, the correct answer is:
A. [tex]\(1 \text{ kJ} \times \frac{1}{\Delta H_{\text{vap}}} \times \text{g/mol liquid}\)[/tex]
