For benzene,
C6H6, the heat of fusion at
its normal melting point of 6 °C is
10.0 kJ/mol.
The entropy change when 1.86 moles of liquid
C6H6 freezes at 6
°C, 1 atm is
? J/K.
Given,
Temperature, T = 6 o C = 279 K
The heat of fusion = 10 KJ/mol
We know,
The change in entropy, S = Heat of fusion / Temperature
= (10 KJ/mol) / 279 K
= (10 * 103 J/mol) / 279 K
= 35.84 J/mol K
Now, for 1.86 moles, the change in entropy, (S)1.86 = 1.86 mol * 35.84 J/mol K
= 66.66 J/K
For benzene, C6H6, the heat of fusion at its normal melting point of 6 °C is...
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