Question

The degree of dissociation of monobasic acid is 0 3 By what percent is the observed depression in freezing point greater than the calculated depression in freezing point ?

Answer 1

Key Equations: 

The observed depression in freezing point (\( \Delta T_f(\text{obs}) \)) is given by:

\[ \Delta T_f(\text{obs}) = K_f \times m \times i \]

The calculated depression in freezing point (\( \Delta T_f(\text{calc}) \)) is given by:

\[ \Delta T_f(\text{calc}) = K_f \times m \]

Where:

• \( K_f \): Freezing point depression constant
• \( m \): Molality of the solution
• \( i \): van’t Hoff factor (for a monobasic acid, \( i = 2 \) assuming complete dissociation)

Percent Difference Formula:

The percent difference is calculated as:

\[ \% \text{difference} = \left[ \frac{\Delta T_f(\text{obs}) - \Delta T_f(\text{calc})}{\Delta T_f(\text{calc})} \right] \times 100 \]

Substitute the equations for \( \Delta T_f(\text{obs}) \) and \( \Delta T_f(\text{calc}) \):

\[ \% \text{difference} = \left[ \frac{K_f \times m \times i - K_f \times m}{K_f \times m} \right] \times 100 \]

Factor out \( K_f \times m \):

\[ \% \text{difference} = \left[ \frac{i - 1}{1} \right] \times 100 \]

Substitute \( i = 2 \):

\[ \% \text{difference} = \left[ \frac{2 - 1}{1} \right] \times 100 = 100\% \]

Adjusted Calculation:

Given the degree of dissociation is 0.3, only 30% of the acid molecules dissociate into ions, leading to an effective \( i < 2 \). This results in a smaller observed depression in freezing point than calculated for full dissociation.

Conclusion:

The observed percent difference in freezing point depression is 30%.

Answer 2