Calculate ΔSfus and ΔSvap for HF.

Determine the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure.

I already calculated ΔSfus as 24.103 J/k*mol and ΔSvap as 86.07 J/K*mol....These are correct (I checked them already). I need help with this part of the question:
Determine the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure.

I got an answer of 183.18 J/K and it marked me wrong and said "The value of ΔSfus above is for 1.00 mol of HF. You need to scale that value up for 7.60 mol of HF. Also, consider the sign of your answer. Does freezing result in an increase or decrease in entropy?"

Could you please help me! Thanks

The problem here may be that you need to add a negative sign to your 183.18 AND you need to watch the significant figures. These data bases are notorious for counting off for the wrong number of s.f.

Yeah but it also said that I need to calculate for 7.60 mol of HF. The answer I got, 183.18 is only for 1.00 mole of HF. It says I need to scale that value up for 7.60 mol of HF. How would I do that????

Didn't you scale it up? You had 24.103 J/mol and you multiplied by 7.60 moles. That looks like scaling it up to me.

So the answer would just be -183.18 ?

Could you please verify the correct answer?(:

And also make sure that -183.18 is correct....I have one more attempt thanks!

Hallo

indeed

Sure! To determine the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure, you need to scale up the value of ΔSfus for 1.00 mol of HF to account for the larger amount of substance.

The value of ΔSfus you calculated, 24.103 J/K*mol, is the entropy change when 1.00 mol of HF(l) freezes. To scale this up for 7.60 mol of HF, you can multiply it by the number of moles:

ΔSfus (for 7.60 mol HF) = ΔSfus (for 1.00 mol HF) * 7.60 mol HF

ΔSfus (for 7.60 mol HF) = 24.103 J/K*mol * 7.60 mol HF

ΔSfus (for 7.60 mol HF) = 183.5488 J/K

Rounding to the appropriate number of significant figures, the entropy change when 7.60 mol of HF(l) freezes is 183.55 J/K.

Now, let's consider the sign of your answer. The freezing process involves a decrease in entropy because the molecules in the liquid state are more disordered than in the solid state. Therefore, the entropy change should be negative.

Hence, the correct answer for the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure is:

ΔSfus = -183.55 J/K

Make sure to include the negative sign when providing your answer.