What is the Entropy of Vaporization for Water at 100 Degrees Celsius?

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Homework Statement

"From the boiling point you found of ∆H(vaporization) at a 100 degrees celsius, find ∆S(vaporization) at 100 degrees celsius from the formula given (see eqn.)
This is for water.

Homework Equations

∆H-T∆S<0

The Attempt at a Solution

From the task before, I found ∆H(vap.) at 100 deg. celsius to be 372.1 K. Used the formula, and it gave me;

∆S>∆H/ T. That gave me ∆S>124.8 J/(mol*K)

Doesnt make much sense to me, is this correct or wrong?

 

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*A little bump* :-)

P.s. I meant ∆H(vap.) at 100 deg. celsius to be 46.2 kJ/ mol
Edit:

Tried: ∆H-T∆S<0

Foumd ∆H to be 46200 J/mol, and T=373,15 K. Gives me that;

46200J/mol - 373,15K*∆S<0

123.8 J/(mol*K)>∆S.

Still does not make any sense! Really confused.