Vodka is advertised to be 80 proof, which is 40% by volume of ethanol, C2H5OH. A

Question

Vodka is advertised to be 80 proof, which is 40% by volume of ethanol, C2H5OH. Assuming the density of the solution is 1.0 g/mL and the density of ethanol is 0.789 g/mL. If the solution is at 25oC, what is the vapor pressure of water over the solution?
Please show all work with the correct answer. Thank you!!!!!!! Vodka is advertised to be 80 proof, which is 40% by volume of ethanol, C2H5OH. Assuming the density of the solution is 1.0 g/mL and the density of ethanol is 0.789 g/mL. If the solution is at 25oC, what is the vapor pressure of water over the solution?
Please show all work with the correct answer. Thank you!!!!!!!
Please show all work with the correct answer. Thank you!!!!!!!

Explanation / Answer

Find vapor pressure of solution at T = 25°C

Assume a basis of:

100 mL:

calculate volumes of each species:

100 mL *(40%/100%) = 40 mL= --> alcohol

100 mL *(60%/100%) = 60 mL= --> water

Recall that the density formula is:

D = mass/ Volume

so

mass = D*V

so

mass of alcohol = 0.789 g/mL *40 mL = 31.56 g of alcohol

mass of water = 1.0 g/mL *60 mL= 60 g of water

calculate moles:

MW of ethanol = 46.06844 g/mol

mol of alcohol = mass/MW = 31.56/46.06844 = 0.6850

MW of ethanol = 46.06844 g/mol

mol of water = mass/MW = 60/18.00 = 3.333333

x-water = 3.333333 / (3.333333 +0.6850) = 0.8295

x-ethanol = 1-x-water = 1- 0.8295 = 0.171

Vapor pressure of ethanol ( T= 25°C) = 5.95 kPa

Vapor pressure of water ( T= 25°C) = 3.1690 kPa

Apply colligative porperties

x1*P° + x2*P° = Pvapor

0.171*5.95 kPa + 0.8295*3.1690 kPa = Pmix

Pmix = 3.6461 kPa

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.