Calculate the entropy change, involved in the conversion of one mole of liquid water at 373 K to vapour at the same temperature (Latent heat of vaporization =2.257 kJ/g).A. ${\rm{105}}{\rm{.9J}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\;$ B. ${\rm{107}}{\rm{.9J}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\;$C. ${\rm{108}}{\rm{.9J}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\;$D. ${\rm{109}}{\rm{.9J}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\;$

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Hint: Vaporisation is determined as the transformation of states of matter from the liquid state to the vapour state.
The latent heat vaporization (∆Hvap) or the enthalpy of vaporization is the enthalpy required to transform a liquid into a gas.

Formula Used:
${\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}{\rm{ = }}\frac{{{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}}}{{{{\rm{T}}_{\rm{b}}}}}$
where
${\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}$=entropy change accompanying vapourization process
${\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}$=enthalpy of vaporisation
${{\rm{T}}_{\rm{b}}}$=boiling point

Complete Step by Step Solution:
In the given reaction, one mole of liquid water at 373 K is converted into vapour.
This reaction involves the transformation of liquid water into a vapour state.
This process is called vapourisation.
This reaction occurs as follows: -
${{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)$
To find the entropy change for this reaction, the enthalpy of vaporization needs to be calculated first.
It is given that heat of vaporization or
${{\rm{q}}_{{\rm{vap}}}}$=2.257 kJ/g = 2257 J/g
Mass of 1 mole of water=18g/mol
So, the enthalpy of vaporisation or
${\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}$
=${{\rm{q}}_{{\rm{vap}}}}$×(mass of 1 mole of water)
=2257 J/g(18 g/mol)
=40626 J/mol
So, ${\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}$ is 40626 J/mol
We are given the boiling point,
${{\rm{T}}_{\rm{b}}}$=373K.
So, we have ${\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}$=enthalpy of vaporisation=40626 J/mol
${{\rm{T}}_{\rm{b}}}$=boiling point=373K
${\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}{\rm{ = }}\frac{{{\rm{40626Jmo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{{{\rm{373K}}}}$
${\rm{ = 108}}{\rm{.91Jmo}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}$
So, option C is correct.

Note: While attending to the question, units of temperature, the heat of vaporisation, enthalpy of vaporisation, and entropy must be surely mentioned. Unit of entropy the conclusive statement must be mentioned. The heat of vaporisation is given in kJ/mol while the given options are in J/Kmol. So, the conversion of kJ/mol into J/mol is important.