Problem solutions

1. Calculate the vapor pressure of a mixture containing  252 g of n-pentane (Mw = 72) and 1400 g of n-eptane (Mw = 100) at 20^oC. The vapor pressure of n-pentane and n-eptane are 420 mm Hg and 36 mm Hg respectively.

According to Raoult's law, the vapor pressure exercised by a component of a mixture can be calculated as follows

P = P^o x

where

P is the vapor pressure of the component in the mixture.
P^o is the vapor pressure of the pure component.
x is the molar fraction of the component in the mixture.

Calculation of molar fractions (x)

moles n-pentane = 252/72 = 3.5

moles n-eptano = 1400/100 = 14

Totals = 3.5 + 14 = 17.5 moles
x_n-pentane = 3.5/17.5 = 0.2
x_n-eptane = 14/17.5 = 0.8

P_n-pentane =  0.2 x 420 = 84 mm Hg
P_n-eptane =  0.8 x 36 = 28.8 mm Hg

and the vapor pressure of mixture is

P_mixture = 84 + 28.8 = 112.8 mm

Note: Although the Raoult's law is valid for diluted (ideal) solution, its application to concentrated (non-ideal) solutions does not lead to variation of the order of  magnitude of the calculated vapor pressure.

2. Calculate the boiling point (at 1 atm) of a solution containing 116 g of acetone (Mw = 58) and 72 g of water (Mw = 18) by using the following table:
 

 

Temperature oC	Vapor pressure (atm) Acetone	Vapor pressure (atm) Water
60	1.14	0.198
70	1.58	0.312
80	2.12	0.456
90	2.81	0.694

A liquid starts to boil when its vapor pressure matches the atmospheric pressure (1 atm in this case). Thus, according to Raoult's law

P = x_acetone P^o _acetone + x_water P^o_water = 1 atm

From the given data we can calculate the molar fractions

moles_acetone = 116/58 = 2
moles_water = 72/18 = 4

total moles = 6

 
x_acetone = 2/4 = 1/3

x_water = 4/6 = 2/3

thus

P = 2/3 P^o _acetone + 1/3 P^o_water= 1 atm

By trials, using the table, we can find the values of vapor pressure which satisfies the above equation. The best result is obtained by using  the values at  80^oC :

P = 2/3 0.456 + 1/3 2.12 = 1.01 atm

then the boiling point is about 80^oC.

3. A mixture of water and acetone at 756 mm boils at 70^oC. Calculate the percentage composition of the  mixture using the following table:
 

 

Temperature oC	Vapor pressure (atm) Acetone	Vapor pressure (atm) Water
60	1.14	0.198
70	1.58	0.312
80	2.12	0.456
90	2.81	0.694

According to Raoult's law

P = x_acetone P^o _acetone + x_water P^o_water = 756/760 =0.995 atm

by substituting the values at 70^oC we have

P = x_acetone 1.58 + x_acqua 0.312= 0.995

and remembering that the sum of molar fractions is 1

 

xacetone + xacqua = 1

xacqua = 1 - xacetone

thus

x_acetone 1.58 + 0.312 (1 - x_acetone) = 0.995

x_acqua 1.58 + 0.312 - 0.312 x_acetone = 0.995

x_acetone 1.26+ 0.312 = 0.995

x_acetone = (0.995 - 0.312)/ 1.26 = 0.54

from which

x_water = 1 - 0.54 = 0.46

considering 100 moles of solution we have

54 moles of  acetone corresponding to 54x58 =3132 grams and 46 moles of water corresponding to 46x18 = 828 grams. Eventually,

% water = (828)(100)/3132 + 828) = 21%

% acetone = 100-21 = 79%

4. The addition of 114 grams of sucrose to 1000 grams of water lowers the vapor pressure of water from 17.540 to 17.435. Calculate the molecular weight of sucrose.

The variation in pressure is 17.540 - 17.435 = 0.105 then

0.105 = 17.540 x₂  = n_solute/(n_solvent + n_solute)

n_solvent = 1000/mw _water = 1000/18 = 55.55 moles
0.105 = 17.540 x₂  = 17.540 n_solute/(55.55 + n_solute)

by solving this equation we obtain n_solute = 0.335 moles and thus

MW = 114/0.335 = 340

The calculated molecular weight (340) is very close to the true molecular weight of sucrose (342).

5. Calculate the osmotic pressure at 20^o of a suspension containing 60g/l of solid particles each particle having a mass of 10^-9 grams (1 nanogram).

p = MRT
T = 278 + 20 = 298 ^oK
R = 0.0823

The number of particles in one liter of solution is

60/10^-9 = 6 x 10¹⁰

Remembering that one mole contains an Avogadro number of particles (6 x 10²³), the number of moles in one liter of suspension is

6 x 10¹⁰/6 x 10²³ =  10^-13

which corresponds to the molarity of the suspension. Thus

p = (0.0823)(298) (10^-13) = 24.5 x 10^-13 atm

6. Calculate the osmotic pressure of a solution 0.1 molale of sucrose (Mw = 342) at  20^oC.

To solve the problem we need to convert the molality in molarity

Molality = moles/kilogram of solvent

Molarity = moles/liter of solution

34.2 + 1000 grams of solvent = 1034.2

0.1 moles : 1034.2 = M : 1000 grams of solution

M = 100/1034.2 = 0.097

p = (0.097)(0.0820578 liter atm mol^-1 K^-1)(293) = 2.34 atm