Be sure to answer all parts. The vapor pressure of ethanol (C2H5O11) at 20 degre

Question

Be sure to answer all parts. The vapor pressure of ethanol (C2H5O11) at 20 degree C is 44 mmHg, and the vapor pressure of methanol (CH3O11) at the same temperature is 94 mmHg. A mixture of 27.5 g of methanol and 44.5 g of ethanol is prepared and can be assumed to behave as an ideal solution. Calculate the vapor pressure of methanol and ethanol above this solution at 20 degree C. Be sure to report your answers to the correct number of significant figures. calculate the mole fraction of methanol and ethanol in the vapor above this solution at 20 degree C.

Explanation / Answer

Let us denote ethanol by the subscript A and methanol by the subscript B

Given :

Pure vapor pressure of ethanol, PAo = 44 mm Hg

Pure vapor pressure of methanol, PBo = 94 mm Hg

Moles of ethanol taken, NA = Mass taken / MW = 44.5/46 = 0.967 moles

Moles of methanol taken, NB = Mass taken / MW = 27.5/32 = 0.859 moles

Mole fraction of ethanol, XA = NA/(NA + NB) = 0.53

Mole fraction of methanol, XB = 1 - XA = 0.47

Using Raoult's law of partial pressure,

Vapor pressure of ethanol above solution = PA = PAo*XA = 44*0.53 = 23.32 mm Hg

Vapor pressure of methanol above solution = PB = PBo*XB = 94*0.47 = 44.18 mm Hg

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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