Question A2 Total for Question A2: 12 marks] 124.4 g of ethanol (CH3CH2OH and 34

Question

Question A2 Total for Question A2: 12 marks] 124.4 g of ethanol (CH3CH2OH and 34.3 g of isooctane (CsHi8) are mixed to form a solution at 40 °C. The vapour pressure of pure ethanol is 130.0 torr and of pure isooctane is 43.9 torr at 40 °C respectively. (a)Assuming ideal behaviour, calculate the total vapour pressure above this solution. 13 marks] (b) Calculate the mole fraction of ethanol and isooctane in the vapour above the solution [2 marks] (c) The experimentally determined vapour pressure above the solution is 185.9 torr and the mole fraction of ethanol in the vapour is 0.6667. Calculate the activities and 4 marks] State whether this solution exhibits positive or negative deviation from ideal activity coefficients for ethanol and isooctane. (d) behaviour. Explain your decision in terms of the relative strength of molecular interactions based on the structure of each molecule. 13 marks]

Explanation / Answer

a) Ethanol Molecular weight=> 2* molar mass of carbon+ 6* molar mass of Hydrogen +1*molar mass of Oxygen

=> 2*12+6*1+1*16=> 46 g/mol

so, number of moles of ethanol=> 124.4g/46 g/mol=> 2.70 moles

molecular weight of isooctane=> 8*molar mass of Carbon+18*molar mass of hydrogen

=> 8*12+18*1=> 96+18=> 114 g/mol

so, number of moles of isooctane=> 34.3 g/114g/mol=> 0.30 moles

now, mole fraction of ethanol=> 2.70mol/(2.70+0.30)mol=> 0.9

mole fraction of isooctane=> 1-0.9=> 0.1

so, assuming ideal behavior, total vapor pressure above the solution=> 0.9*130.0 torr+ 0.1* 43.9 torr=> 121.39 torr

b) Partial vapour pressure of ethanol in vapor phase=> 0.9*130.0 torr=> 117 torr

Partial pressure of isooctane in vapor phase=> 0.1* 43.9 torr=> 4.39 torr

Mole fraction of ethanol in vapour=> ppethanol/(ppethanol+ p isooctane)

=> 117/(117+4.39)=> 0.96

similarly for isooctane=> 0.04

c) Experimentally determined vapor pressure=> 185.9 torr

mole fraction of ethanol in vapor=> 0.67

so, its partial pressure=> 0.67*185.9 torr=> 124.55 torr

so, activity=> 124.55torr/130 torr=> 0.96

now, its activity coefficient => actual vapor pressure/theoretical raoults vapor pressure=> 124.55/117=> 1.06

mole fraction of isooctane= 0.33

so, its partial pressure=> 0.33* 185.9=> 61.34 torr

so, activity=> 61.34 torr/ 43.9 torr=> 1.39

Activity coefficient => actual V.P/theoretical V.P.=> 61.34/4.39= 13.97

d) We can see that the actual vapor pressure is greater than the theoretical vapour pressure. Means we see a positive deviation from raoult's law. This is due to repulsion among the molecules in solution due to which they escape more than expected in ideal solution and thus have a greater vapor pressure.

Isooctane has very repelling interaction hence its activity is very high.

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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