Acetic acid (CH3COOH) has a heat of vaporization of 23.4 kJ/mol. What heat is required to vaporize 250 g of acetic acid?

97.5 kJ of heat is required to vaporize 250 g of acetic acid

Explanation:

Acetic acid → CH₃COOH

The heat of vaporization is 23.4 kJ/mol, which it means that 1 mol of acetic releases 23.4 kJ of energy.

Let's convert our mass of acetic to moles:

250 g . 1 mol /60 g = 4.17 moles

Now we propose this rule of three:

1 mol of acetic acid releases 23.4 kJ

Then, 4.17 moles will release (4.17 . 23.4) / 1 = 97.5 kJ

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%