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.  

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.