6.24 g of ethanol are vaporized by supplying 5.89 KJ of heat energy . What is the ebnthalpy of vapourisation of ethanol?

Options

A. 43.42 KJ
B. 47.0 KJ
C. 21.75 KJ
D. 435.0 KJ

Answer

To find the enthalpy of vaporization of ethanol, we can follow these steps:

### Step 1: Determine the molar mass of ethanol (C2H5OH)
The molar mass of ethanol can be calculated as follows:
- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol
- Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol

Adding these together:
\[ \text{Molar mass of ethanol} = 24.02 + 6.048 + 16.00 = 46.068 \, \text{g/mol} \]
For practical purposes, we can round this to 46 g/mol.

### Step 2: Calculate the number of moles of ethanol
Using the formula for calculating moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{6.24 \, \text{g}}{46 \, \text{g/mol}} \approx 0.1357 \, \text{moles} \]

### Step 3: Calculate the enthalpy of vaporization
The enthalpy of vaporization (ΔH) can be calculated using the formula:
\[ \Delta H = \frac{\text{Heat supplied}}{\text{Number of moles}} \]
Substituting the values:
\[ \Delta H = \frac{5.89 \, \text{kJ}}{0.1357 \, \text{moles}} \approx 43.42 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of vaporization of ethanol is approximately **43.42 kJ/mol**.

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