Abnormal Molar Masses

Since colligative properties depend upon the number of particles of the solute, in some cases where the solute associates or dissociates in solution, abnormal results for molecular masses are obtained.

Van't Hoff Factor: Van't Hoff, in order to account for all abnormal cases introduced a factor i known as the Van't Hoff factor, such that

$i=\frac{Observedcolligativeproperty(actual)}{Theoreticalcolligativeproperty(expected)}=\frac{Theorticalmolecularmass}{observedmolecularmass}$

$=\frac{No.ofmoleculesactuallypresent}{No.ofmolecules\exp ectedtobepresent}$

i $>$ 1 for dissociation of solute in solution

i $<$ 1 for association of solute.

i = 1 when solute neither associate non dissociate.

Illustration 1.

The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is

(A) 60% (B) 83.5%

(C) 46.7% (D) 60.23%

Solution: i for AgNO₃ = $\frac{Normal\text{ mol}\text{.wt}}{Observed\text{ mol}\text{.wt}\text{.}}$ = 1 + a

a = $\frac{170}{92.64}$ – 1 = 0.835 = 83.5 %

Hence, (B) is correct.

Illustration 2.

20 g of a binary electrolyte (mol.wt. = 100) are dissolved in 500 g of water. The freezing of the solution is –0.74°C, K_f = 1.86 K. molality^–1. The degree of ionisation of the electrolyte is

(A) 50% (B) 75%

(C) 100% (D) 0

Solution: $\Delta$T_f = $\frac{1000\times K_f\times w}{M\times W}$

$\therefore$ 0.74 = $\frac{100\times 1.86\times 20}{M\times 500}$ $\Rightarrow$ M = 100

Now, $\frac{Normal\text{ mol}\text{.wt}}{\exp .\text{ mol}\text{.wt}\text{.}}$= 1 +$\alpha$ = $\frac{100}{100}$

So, $\alpha$ = 0

Hence, (D) is correct.

Illustration 3.

Acetic acid (CH₃COOH) associates in benzene to form double molecules. 1.65 g of acetic acid when dissolved in 100g of benzene raised the boiling point by 0.36°C. Calculate the Van't Hoff Factor and the degree of association of acetic acid in benzene (Molal elevation constant of benzene is 2.57).

Solution: Normal molar mass of acetic acid = 60

Observed molar mass of acetic acid

$M=K_b\frac{w_{solute}}{W_{solvent}^{}(ingms)\Delta T}\times 1000$=$\frac{2.57\times 1.65\times 1000}{100\times 0.36}=118$

Van't Hoff Factor = $\frac{Normalmolarmass}{Observedmolarmass}=\frac{60g{\mathrm{mol}}^{-1}}{118{\mathrm{gmol}}^{-1}}$

= 0.508

0.508 = $1-\left[\alpha \left(1-\frac{1}{n}\right)\right]=1-\left[\alpha \left(1-\frac{1}{2}\right)\right]$ = 1 – $\alpha$/2

$\alpha$/2 = 1– 0.508 = 0.492

$\therefore$ $\alpha$ = 2 × 0.492 = 0.984

Thus acetic acid is 98.3% associated in benzene.

Illustration 4.

River water is found to contain 11.7% NaCl, 9.5% MgCl₂, and 8.4%. NaHCO₃ by weight of solution. Calculate its normal boiling point assuming 90% ionization of NaCl,70% ionization of MgCl₂ and 50% ionization of NaHCO₃ (K_b for water = 0.52 k kg mol^-1)

Solution: n_NaCl = $\frac{11.7}{58.5}$ = 0.2

$n_{MgC{l}_2}$ = $\frac{9.5}{95}$ = 0.1

$\frac{11.7}{58.5}$= $\frac{8.4}{84}$ = 0.1

i_NaCl = 1+ $\alpha$ = 1+ 0.9 = 1.9

$i_{MgC{l}_2}$ = 1 + 2$\alpha$ = 1+ 0.7 ´ 2 = 2.4

$i_{NaHC{O}_3}$ = 1+ 2$\alpha$ = 1+ 0.5 ´ 2 = 2.0

Weight of solvent = 100 – (11.7 + 9.5 + 8.4) = 70.4 g

$\Delta$T_b = $\frac{(i_{NaCl}\times n_{NaCl}+i_{MgC{l}_2}\times n_{MgC{l}_2}+i_{NaHC{O}_3}\times n_{NaHC{O}_3})\times K_b\times 1000}{Weight\text{ of solvent}}$

= $\frac{(1.9\times 0.2+2.4\times 0.1+2\times 0.1)\times 0.52\times 1000}{70.4}$

= 6.0657°C

$\therefore$ Boiling point of solution = 100 + 5.94 = 106.057°C