Based on the following thermodynamic data, calculate the boiling pressure of ethanol in degrees celcius

Substance Delta Hf ( Kj/mol) Delta Sf ( J/K*mol)

C2H5OH (L) --- -277.7 160.6

C2H5OH (G) --- -235.1 282.6

76.03

Explanation:

We have the equation

T= Delta H/ Delta S

For this we need to find the Delta H of vaporization for the boiling point. We can obtain this by taking delta H(products) - delta H(reactants). In this case we will take (-235.1 kj/mol) - (-277.1 kj/mol) = 42.6 kj/mol.

We will then need the delta S of vaporization, which can be obtained by taking delta S(products) - delta S(reactants). Here we will take (282.6 j/k*mol) - (160.6 j/k*mol) = 122 j/k*mol. We still need to convert this to kj/mol so we divide 122 by 1000 to get 0.122 kj/mol as delta S of vaporization.

Then we plug in these values into the equation mentioned above as follows.

T= 42.6/0.122

T= 349.18 K

All we need to do now is to convert this into degrees Celsius by taking 349.18 - 273.15 to end up with an answer of 76.03 degrees Celsius.

the answer is the first option

explanation:

just googled it m8. not that hard

lol

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%