Question

1.8 g water is vapourised by supplying 4 kJ heat at \(100^\circ\text{C}\). What is the heat of vapourisation of water at same temperature?

• A. \(8 \text{ kJ mol}^{-1}\)
• B. \(40 \text{ kJ mol}^{-1}\)
• C. \(18 \text{ kJ mol}^{-1}\)
• D. \(32 \text{ kJ mol}^{-1}\)

Hint:

To find heat of vapourisation: \[ \Delta H_{\text{vap}}=\frac{\text{heat supplied}}{\text{number of moles}} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The heat of vaporization (or enthalpy of vaporization, \(\Delta H_{\text{vap}}\)) is defined as the amount of heat energy required to vaporize exactly one mole of a liquid substance at its boiling point under standard pressure.
The problem gives us the heat required for a specific mass, and we must convert this to a per-mole basis.
Step 2: Key Formula or Approach:
First, calculate the number of moles of water using \(n = \frac{\text{Mass}}{\text{Molar Mass}}\).
Then find the heat of vaporization per mole using \(\Delta H_{\text{vap}} = \frac{q}{n}\), where \(q\) is the total heat supplied.
Step 3: Detailed Explanation:
The given mass of water (\(\text{H}_2\text{O}\)) is \(m = 1.8 \text{ g}\).
The heat energy supplied is \(q = 4 \text{ kJ}\).
The chemical formula of water is \(\text{H}_2\text{O}\). Its molar mass is \((2 \times 1) + 16 = 18 \text{ g/mol}\).
Let's calculate the number of moles of water in the \(1.8 \text{ g}\) sample:
\[ \text{Moles of } \text{H}_2\text{O} (n) = \frac{1.8 \text{ g}}{18 \text{ g/mol}} = 0.1 \text{ moles} \] The problem states that vaporizing \(0.1 \text{ moles}\) of water requires \(4 \text{ kJ}\) of heat.
To find the heat of vaporization, we need to determine the heat required for \(1 \text{ entire mole}\).
We set up a simple ratio:
\[ \Delta H_{\text{vap}} = \frac{\text{Heat supplied}}{\text{Moles vaporized}} \] \[ \Delta H_{\text{vap}} = \frac{4 \text{ kJ}}{0.1 \text{ mol}} \] \[ \Delta H_{\text{vap}} = 40 \text{ kJ/mol} \] Step 4: Final Answer:
The heat of vaporization of water is \(40 \text{ kJ mol}^{-1}\).