Chemistry Abnormal Molar Masses :
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### Topics Covered :

● Abnormal Molar Mass
● Abnormal Molar Mass (Dissociation)
● Abnormal Molar Mass (Association)

### Abnormal Molar Masses

color{green}("For Dissociation :") We know that ionic compounds when dissolved in water dissociate into cations and anions.

● For example, if we dissolve one mole of KCl (74.5 g) in water, we expect one mole each of K^+ and Cl^- ions to be released in the solution. If this happens, there would be two moles of particles in the solution. So, one mole of KCl in one kg of water would be expected to increase the boiling point by 2xx0.52K = 1.04 K.

● If we did not know about the degree of dissociation, we could be led to conclude that the mass of 2 mol particles is 74.5 g and the mass of one mole of KCl would be 37.25 g. This conclude that, when there is dissociation of solute into ions, the experimentally determined molar mass is always lower than the true value.

color{green}("For Association :")

● Molecules of ethanoic acid (acetic acid) dimerise in benzene due to hydrogen bonding. This normally happens in solvents of low dielectric constant. In this case the number of particles is reduced due to dimerisation. Association of molecules is depicted as follows :

● If all the molecules of ethanoic acid associate in benzene, then ΔT_b or ΔT_f for ethanoic acid will be half of the normal value.

● Therefore, the molar mass calculated on the basis of this ΔT_b or ΔT_f will be twice the expected value.

color{green}("Abnormal Mass") : Such a molar mass that is either lower or higher than the expected or normal value is called as abnormal molar mass.

In 1880 van’t Hoff introduced a factor i, known as the van’t Hoff factor, to account for the extent of dissociation or association. This factor i is defined as :

color{green}(i =text{Normal molar mass}/text{Abnormal molar mass} )

 = color{green}(text(Observed colligative property)/text(Calculated colligative property))

color{green}(i = text(Total number of moles of particles after association/dissociation)/text(Number of moles of particles before association/dissociation))

Here, abnormal molar mass = experimentally determined molar mass

Calculated colligative properties are obtained by assuming that the non-volatile solute is neither associated nor dissociated.

● In case of association, value of i is less than unity

● In case of dissociation, it is greater than unity.

=> For example, the value of i for aqueous KCl solution is close to 2, while the value for ethanoic acid in benzene is nearly 0.5.

=> Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows :

Relative lowering of vapour pressure of solvent. (p_1^0 - p_1)/(p_1^0) = i n_2/n_1

Elevation of Boiling point, DeltaT_b = i K_b m

Depression of Freezing point, DeltaT_f = i K_f m

Osmotic pressure of solution, pi = i n_2RT//V

Table 2.4 depicts values of the factor, i for several strong electrolytes. For KCl, NaCl and MgSO_4, i approach 2 as the solution becomes very dilute. As expected, the value of i gets close to 3 for K_2SO_4.
Q 2966856775

2 g of benzoic acid (C_6H_5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol^(–1). What is the percentage association of acid if it forms dimer in solution?

Solution:

The given quantities are: w_2 = 2g ; K_f = 4.9 K kg mol^(-1) ; w_1 = 25 g

Substituting these values in equation (2.36) we get: M_2 = (4.9 K kg mol^(-1) xx 2g xx 1000g kg^(-1))/(25 g xx 1.62 K) = 241.98 g mol^(-1)

Thus, experimental molar mass of benzoic acid in benzene is  = 241.98 g mol^(-1)

2C_6H_5 COOH ⇌ ( C_6H_5 COOH)_2

If x represents the degree of association of the solute then we would have (1 – x ) mol of benzoic acid left in unassociated form and correspondingly x/2 as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is: 1-x +x/2 = 1-x/2

Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i.

But i = text(Normal molar mass)/text(Abnormal molar mass)

 = (122 g mol^(-1))/(241.98 g mol^(-1))

or x/2 = 1-(122)/(241.98) = 1-0.504 = 0.496

or x = 2xx0.496 = 0.992

Therefore, degree of association of benzoic acid in benzene is 99.2 %.
Q 2926056871

0.6 mL of acetic acid (CH_3COOH), having density 1.06 g mL^(–1), is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the dissociation constant of acid.

Solution:

Number of moles of acetic acid  = (text(0.6 mL xx 1.06 g mL^(-1))/(60g mol^(-1))

 = 0.0106 mol = n

Molality = (0.0106 mol)/(100 mL xx 1 g mL^(-1)) = 0.0106 mol kg^(-1) Using equation (2.35)

DeltaT_f = 1.86 K kg mol^(-1) xx 0.0106 mol kg^(-1) = 0.0197 K

van’t Hoff Factor (i) = text(Observed freezing point)/text(Calculated freezing point) = (0.0205 K)/(0.0197 K) = 1.041

Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen ions per molecule of acetic acid. If x is the degree of dissociation of acetic acid, then we would have n (1 – x) moles of undissociated acetic acid, nx moles of CH_3COO^– and nx
moles of H^+ ions,

underset( n mol)(CH_3COOH) ⇌ underset(0)(H^+) + underset(0)(CH_3COO^-)

n(1-x) qquad qquad nx mol qquad qquad nx mol

Thus total moles of particles are: n (1-x +x+x) = n (1+x)

i = ( n(1+x))/n = 1+x = 1.041

Thus degree of dissociation of acetic acid  = x = 1.041 - 1.000 = 0.041

Then [CH_3COOH] = n (1-x) 0.0106(1-0.041)

[CH_3COO^-] = nx = 0.0106 xx 0.041 , H^+ = nx = 0.0106 xx 0.041

K_a = ([CH_3COO^-][H^+])/([CH_3COOH]) = (0.0106 xx 0.041xx0.0106xx0.041)/(0.0106(1.00-0.041)

 = 1.86xx10^(-5)