Solutions | Raoult's law | Solved problems

QUESTION 1:

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

SOLUTION:  

From the formulae,  

Molality = moles/kilogram of solvent

Molarity = moles/liter of solution

The grams of sucrose in the solutions are 34.2 and the total weight of the solution is

34.2 + 1000 grams of solvent = 1034.2

Assuming a density is 1 

          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


QUESTION 2:

A mixture of water and acetone at 756 mm boils at 70oC. 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 

SOLUTION:  

According to Raoult's law, 

P = xacetone Po acetone + xwater Powater 

    = 756/760 =0.995 atm

Substituting the values at 70oC ,  

P = xacetone 1.58 + xacqua 0.312 = 0.995

We also know that, 

xacetone + xacqua = 1 

So, xacqua = 1 - xacetone

  So,  xacetone 1.58 + 0.312 (1 - xacetone) = 0.995

xacqua 1.58 + 0.312 - 0.312 xacetone = 0.995

xacetone 1.26+ 0.312 = 0.995

xacetone = (0.995 - 0.312) / 1.26 = 0.54

So,  xwater = 1 - 0.54 = 0.46

Considering that we have 100 moles of solution, 54 moles of  acetone and 46 moles of water is present in the sample. 

54 moles of  acetone is 54 x 58 =3132 grams

and 46 moles of water is 46 x 18 = 828 grams

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

       % acetone = 100-21 = 79%


QUESTION 3:

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.

SOLUTION:  

The difference in pressure: 17.540 - 17.435 = 0.105 

Then, 

0.105 = 17.540 x= nsolute/(nsolvent + nsolute)

nsolvent = 1000/mw water = 1000/18 = 55.55 moles

0.105 = 17.540 x= 17.540 nsolute/(55.55 + nsolute)

by solving this,

         nsolute = 0.335 moles and 

MW = 114/0.335 = 340

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


QUESTION 4:

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

SOLUTION:   

Formula:  p = MRT ( T = 278 + 20 = 298 oK and R = 0.0823)

The number of particles in one liter of solution is = 60/10-9 = 6 x 1010

1 mole of any substance contains an Avogadro number of particles (6 x 1023)

The number of moles in one liter of suspension = 6 x 1010/6 x 1023 =  10-13

which is equal to the molarity of the suspension.

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


QUESTION 5:

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

SOLUTION: 

From  Raoult's law, P = Po x

Here,

P = vapor pressure of the component in the mixture.

Po= vapor pressure of the pure component.

x = Molar fraction of the component in the mixture.

Calculation of mole fraction: 

Mole fraction = moles of one component / Total moles in solution

Moles n-pentane = 252/72 = 3.5

Moles n-eptano = 1400/100 = 14

Totals = 3.5 + 14 = 17.5 moles

xn-pentane = 3.5/17.5 = 0.2 ,  xn-eptane = 14/17.5 = 0.8

So,  Pn-pentane =  0.2 x 420 = 84 mm Hg ; Pn-eptane =  0.8 x 36 = 28.8 mm Hg

and the vapor pressure of mixture is, Pmixture = 84 + 28.8 = 112.8 mm



QUESTION 6:

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 

SOLUTION: 

A liquid starts to boil when its vapor pressure matches the atmospheric pressure (1 atm in this case).

According to Raoult's law

P = xacetone Po acetone + xwater Powater = 1 atm

Calculate the molar fractions: 

Mole fraction = moles of one component / Total moles in solution

molesacetone = 116/58 = 2 ; moleswater = 72/18 = 4

So, Total moles = 2 + 4 =  6 

Substituting the given values, 

xacetone = 2/4 = 1/3  ;  xwater = 4/6 = 2/3

So,  P = 2/3 Po acetone + 1/3 Powater= 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  80oC :

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

So, the boiling point is about 80oC.

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