Question

The enthalpy of fusion of water is 1.435 kcal/mol.
The molar entropy change for the melting of ice at
0°C is
[AIPMT (Prelims)-2012]
(1) 5.260 cal/mol K)
(2) 0.526 cal/(mol K)
(3) 10.52 cal/mol K)
(4) 21.04 cal/(mol K)​

Answer 1

Answer:

Answer:

Formula: ΔH = TΔS

Where, ΔH symbolizes Enthalpy of Fusion, T represents Temperature in Kelvin and ΔS represent Entropy.

According to the question,

→ Enthalpy of Fusion = ΔH = 1.435 Kcal/mol

→ Enthalpy of Fusion = ΔH = 1435 cal/mol

It is given melting point of ice is 0°C. Converting it to Kelvin we get:

→ Melting Point of Ice = 0°C + 273 = 273 K

Now we are required to find the value of Molar Entropy ( ΔS )

Since ΔH is in terms of Calories per mole, we can directly substitute it in the formula.

→ 1435 cal/mol = 273 K ×ΔS

→ ΔS = 1435 cal/mole / 273 K

→ ΔS = 5.256 cal/mole.K

→ ΔS ≈ 5.26 cal/mole.K

Hence Option (1) is the correct answer.