Question

If 1 gm of solute dimerise upto 75% in 100 gm of H₂O & show depression in freezing point 0.093°C. Find average molecular weight of solute in the solution ($K_f(H_2O)$ = 1.86 K-kg $mol^{-1}$).

Answer 1

200

Answer 2

The problem involves the depression in freezing point of a solution where the solute undergoes dimerization. The depression in freezing point ($\Delta T_f$) is a colligative property and is related to the molality of the solution by the formula:

$\Delta T_f = i \cdot K_f \cdot m$

where $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant of the solvent, and $m$ is the molality of the solution calculated based on the initial amount of solute.

The solute undergoes dimerization, which is an association reaction where two monomer molecules combine to form one dimer molecule: $2A \rightleftharpoons A_2$.

Let the initial number of moles of the solute be $n_0$. If a fraction $\alpha$ of the solute molecules dimerizes, the moles of monomer that react are $\alpha n_0$. These reacting moles form $\alpha n_0 / 2$ moles of dimer.

The number of moles of monomer remaining is $n_0 - \alpha n_0 = n_0(1 - \alpha)$.

The total number of moles of particles in the solution at equilibrium is the sum of the moles of remaining monomer and the moles of dimer formed:

$n_{total} = n_0(1 - \alpha) + \frac{\alpha n_0}{2} = n_0 \left(1 - \alpha + \frac{\alpha}{2}\right) = n_0 \left(1 - \frac{\alpha}{2}\right)$.

The van't Hoff factor ($i$) is the ratio of the total number of moles of particles in the solution after association/dissociation to the initial number of moles of solute:

$i = \frac{n_{total}}{n_0} = \frac{n_0(1 - \alpha/2)}{n_0} = 1 - \frac{\alpha}{2}$.

We are given that 75% of the solute dimerizes, which means the fraction of dimerization $\alpha = 0.75$.

Substituting this value into the formula for $i$:

$i = 1 - \frac{0.75}{2} = 1 - 0.375 = 0.625$.

The molality ($m$) in the formula $\Delta T_f = i \cdot K_f \cdot m$ is the ideal molality, calculated using the initial number of moles of solute ($n_0$) and the mass of the solvent.

Mass of solute = 1 gm.

Mass of solvent (H₂O) = 100 gm = 0.1 kg.

Let the normal molar mass of the solute (monomer) be $M$.

The initial number of moles of solute is $n_0 = \frac{\text{Mass of solute}}{M} = \frac{1}{M}$ moles.

The molality based on the initial amount of solute is $m = \frac{n_0}{\text{Mass of solvent in kg}} = \frac{1/M}{0.1} = \frac{1}{0.1M} = \frac{10}{M}$.

Now substitute the given values into the freezing point depression formula:

$\Delta T_f = 0.093$ °C

$K_f = 1.86$ K-kg/mol (which is the same as °C-kg/mol)

$i = 0.625$

$m = \frac{10}{M}$

$0.093 = 0.625 \cdot 1.86 \cdot \frac{10}{M}$

$0.093 = 1.1625 \cdot \frac{10}{M}$

$0.093 = \frac{11.625}{M}$

$M = \frac{11.625}{0.093} = \frac{11625}{93} = 125$.

The normal molar mass of the solute (monomer) is 125 g/mol.

The question asks for the "average molecular weight of solute in the solution". This refers to the apparent molar mass ($M_{app}$) of the solute as it exists in the solution, considering the equilibrium between monomers and dimers. The apparent molar mass is related to the normal molar mass ($M$) and the van't Hoff factor ($i$) by the relationship:

$i = \frac{\text{Normal molar mass}}{\text{Observed (Average/Apparent) molar mass}} = \frac{M}{M_{app}}$.

Using the values $i = 0.625$ and $M = 125$ g/mol:

$0.625 = \frac{125}{M_{app}}$

$M_{app} = \frac{125}{0.625} = \frac{125}{5/8} = 125 \times \frac{8}{5} = 25 \times 8 = 200$.

The average molecular weight of the solute in the solution is 200 g/mol.

Alternatively, we can first calculate the molality based on the total number of particles in the solution ($m_{actual}$) using the direct relationship $\Delta T_f = K_f \cdot m_{actual}$.

$m_{actual} = \frac{\Delta T_f}{K_f} = \frac{0.093}{1.86} = \frac{93}{1860} = \frac{1}{20} = 0.05$ mol/kg.

This $m_{actual}$ is the total number of moles of particles per kg of solvent.

Total moles of particles in the solution ($n_{total}$) = $m_{actual} \times \text{Mass of solvent in kg} = 0.05 \text{ mol/kg} \times 0.1 \text{ kg} = 0.005$ moles.

The average molecular weight ($M_{app}$) is the mass of the solute divided by the total number of moles of particles in the solution:

$M_{app} = \frac{\text{Mass of solute}}{n_{total}} = \frac{1 \text{ gm}}{0.005 \text{ moles}} = \frac{1}{5/1000} = \frac{1000}{5} = 200$ g/mol.

Both methods yield the same result.