The melting point of water is 0°C at 1 atm pressure because under these conditions: A. ΔS for the process H2O(s) → H2O(l) is positive.
B. ΔS and ΔSsurr for the process H2O(s) → H2O(l) are both positive.
C. ΔS and ΔSsurr for the process H2O(s) → H2O(l) are equal in magnitude and opposite in sign.
D. ΔG is positive for the process H2O(s) → H2O(l).
E. None of these is correct.

The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.

Explanation:

The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:  

H2O (s) ⇒ H2O (l)

The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:  

Entropy change (ΔSsys) = ΔSproduct - ΔSreactant

= (69.9 - 47.89) J mol/K

= 22.0 J mol/K

Therefore, the value of entropy change is positive.  

Now the value of entropy for surrounding ΔSsurr will be,  

ΔSsurr = -ΔHfusion/T  

= -6012 j/mol/273

= -22.0 J/molK

Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.  

according to kinetic-molecular theory, the following would not be considered an ideal gas:

a gas at very low volumes, when gas particles are very close together

a gas at very low temperatures, when gas particles have very little kinetic energy a gas

a gas with highly polar molecules that have very strong intermolecular forces

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