The normal boiling point of water is 100.0 °c and its molar enthalpy of vaporization is 40.67 kj/mol. what is the change in entropy in the system in j/k when 51.1 grams of steam at 1 atm condenses to a liquid at the normal boiling point?

We are going to use this formula:

Δ H = T * ΔS

when Δ H is the change in enthalpy 

and T is the temperature in Kelvin = 100+273= 373 K

and ΔS is the change in entropy

but first, we need to get the value of ΔH:

ΔH = mass * molar mass per mole * enthalpy of vaporization

by substitution:

ΔH = 51.1 g * (1 mole/18g) * 40.67KJ/mole

      = -115.5 KJ

we use negative sign because this is an exothermic reaction.

by substitution on the first formula, we will get the change of entropyΔS:

ΔS = ΔH / T

     = -115.5 KJ / 373K

     = 310 J/K

Answer) by studying proton collisions hope this : )

257 grams h2o i think