Question

Calculate the enthalpy of vaporisation of ethanol if 11.5 g of ethanol is completely vaporised by supplying 11.8 kJ of heat.

• A. $21.7 \text{ kJ mol}^{-1}$
• B. $47.2 \text{ kJ mol}^{-1}$
• C. $65.1 \text{ kJ mol}^{-1}$
• D. $39.0 \text{ kJ mol}^{-1}$

Hint:

$\Delta H$ is always "per mole." Always find the moles of the substance first. Dividing by 0.25 is the same as multiplying by 4!

Answer 1

The Correct Option is B

Step 1: Concept
Enthalpy of vaporisation ($\Delta_{vap}H$) is the heat required to vaporise one mole of a liquid at constant pressure.

Step 2: Meaning
Formula: $\Delta_{vap}H = Q / n$, where $n$ is the number of moles. Molar mass of Ethanol ($\text{C}_2\text{H}_5\text{OH}$) is $46 \text{ g/mol}$.

Step 3: Analysis
- Number of moles ($n$) = Mass Molar mass = $11.5 / 46 = 0.25 \text{ mol}$. - $\Delta_{vap}H = 11.8 \text{ kJ} / 0.25 \text{ mol}$. - $\Delta_{vap}H = 47.2 \text{ kJ/mol}$.

Step 4: Conclusion
The enthalpy of vaporisation is $47.2 \text{ kJ mol}^{-1}$. Final Answer: (B)