Question

SHOW ALL WORK

A) What is the normal vapor pressure of isopropyl alcohol at 25C? (BP = 82.3 C H_vap = 39.9 kJ/mol)

B) What is the normal vapor pressure of water at 25 C?

C) If you make a solotion of 0.1M isopropyl alcohol in water, What will its vapor pressure be?

D) What will the boiling point be?

Answer 1

a)

normal vapor pressure --> P1 = 760, T1 = 82.3°C; P2 = x; T2 = 25°C

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Change negative signs

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

ln(P2/760) = 39900/8.314*(1/(82.3+273.15) - 1/(295))

P2 = 760*exp(39900/8.314*(1/(82.3+273.15) - 1/(295)))

P2 = 47.781 mm Hg

b)

for water...

Pvapor = 23.8 mm Hg

c)

if 0.1 M solution, find vapor pressure

assume 1 liter solution, 0.1 mol of alcohol

1 liter = 1000 g; 0.1  

mol of water = 1000/18 = 55.55 mol of water

dP = xsolute*Psolute

dP = 0.1 / (0.1 +55.55) * 23.8

dP = 0.04276 mm Hg

Psolution = 23.8 - 0.04276 = 23.757 mm Hg

d)

BP -->

Apply Colligative properties

This is a typical example of colligative properties.

Recall that a solute ( non volatile ) can make a depression/increase in the freezing/boiling point via:

dTb = Kb*molality * i

where:

Kb = boiling point constant for the SOLVENT;

molality = moles of SOLUTE / kg of SOLVENT

i = vant hoff coefficient, typically the total ion/molecular concentration.

At the end:

Tb mix = Tb solvent + dTb

Txmix = 100 + 0.512 * 0.1 / 1

Tmix = 100.0512 °C