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

Concept:
Heat of vapourisation is the amount of heat required to convert one mole of liquid into vapour at constant temperature.

Step 1: Find the number of moles of water.
Molar mass of water is: \[ 18\ \text{g mol}^{-1} \] So, \[ n=\frac{1.8}{18}=0.1\ \text{mol} \]

Step 2: Use the heat given for 0.1 mol water.
Given heat required: \[ q = 4\ \text{kJ} \] This heat is for: \[ 0.1\ \text{mol} \]

Step 3: Calculate heat of vapourisation per mole.
\[ \Delta H_{\text{vap}}=\frac{4}{0.1}=40\ \text{kJ mol}^{-1} \] Hence, the correct answer is:
\[ \boxed{(B)\ 40\ \text{kJ mol}^{-1}} \]