Question

The latent heat of fusion of ice is 0.333 $KJ{{g}^{-1}}$ . the increase in entropy when 1 mole water melts at 0$^{0}C$ will be:
a.) 21.98 KJ${{K}^{-1}}mo{{l}^{-1}}$
b.) 21.98 Kcal${{K}^{-1}}mo{{l}^{-1}}$
c.) 21.98 J${{K}^{-1}}mo{{l}^{-1}}$
d.) 21.98 cal${{K}^{-1}}mo{{l}^{-1}}$

Answer 1

To solve this question, recall the definition of value of latent heat of fusion for ice. The amount of heat energy which is absorbed or released when a solid changes into liquid state at the atmospheric pressure at its melting point. This is known as latent heat of fusion.

Complete Solution :
Latent heat is the hidden form of energy which means that it is the energy which is used to change its state without increasing the temperature. If we talk about the normal conditions, at atmospheric pressure the value of latent heat of ice is 80cal/g.
Given in the question:
The latent heat of fusion of ice = 0.333 $KJ{{g}^{-1}}$
1 mole of water = 18 g of water = m
$\Delta {{H}_{vap}}$ will be equal to the product of mass of water to the latent heat of fusion of ice.

& \Delta {{H}_{vap}}=ml \\\ & \Delta {{H}_{vap}}=18\times 0.333\times {{10}^{3}} \\\ & =5994J/mol \\\ \end{aligned}$$ Hence, the increase in entropy = $\Delta S = \dfrac{\Delta {{H}_{vap}}}{T}=\dfrac{5994j/mol}{273K} = 21.98J{{K}^{-1}}mo{{l}^{-1}}$ Hence, the correct answer is option (C) i.e. The latent heat of fusion of ice is 0.333 $KJ{{g}^{-1}}$ . the increase in entropy when 1 mole water melts at 0$^{0}C$ will be $21.98J{{K}^{-1}}mo{{l}^{-1}}$. **So, the correct answer is “Option A”.** **Note:** For general use, you should remember the basic value of latent heat of fusion for ice and some other important values. The most important part in such questions is the unit as you can see, the value given in all the options is exactly the same the difference is only in units. So it is most important to remember the unit conversion and to use the correct unit while solving such questions.