Question

The observed molecular weight of a 1:1 strong electrolyte is 117 g mol\(^{-1}\) as determined by the depression of freezing point method. Its theoretical molecular weight is 60 g mol\(^{-1}\). The percentage dissociation of the electrolyte is:

• A. 90%
• B. 95%
• C. 100%
• D. 85%
• E. 80%

Hint:

To calculate the percentage dissociation of an electrolyte, use the formula relating observed and theoretical molecular weights, and solve for the degree of dissociation.

Answer 1

The Correct Option is A

The observed molecular weight \( M_{\text{obs}} \) of an electrolyte can be related to the theoretical molecular weight \( M_{\text{theo}} \) and the degree of dissociation \( \alpha \) using the formula:

\[ M_{\text{obs}} = \frac{M_{\text{theo}}}{1 + \alpha (n - 1)} \] where:

- \( M_{\text{obs}} \) is the observed molecular weight,
- \( M_{\text{theo}} \) is the theoretical molecular weight,
- \( \alpha \) is the degree of dissociation (fraction of the electrolyte that dissociates),
- \( n \) is the number of ions produced per formula unit of electrolyte (for a 1:1 electrolyte, \( n = 2 \)).

Given:
- \( M_{\text{obs}} = 117 \, \text{g/mol} \),
- \( M_{\text{theo}} = 60 \, \text{g/mol} \),
- \( n = 2 \).

Substitute these values into the formula:

\[ 117 = \frac{60}{1 + \alpha (2 - 1)} \] \[ 117 = \frac{60}{1 + \alpha} \] Multiplying both sides by \( (1 + \alpha) \), we get:

\[ 117(1 + \alpha) = 60 \] \[ 117 + 117\alpha = 60 \] \[ 117\alpha = 60 - 117 = -57 \] \[ \alpha = \frac{-57}{117} = 0.486 \] Thus, the degree of dissociation \( \alpha \) is approximately 0.486, which corresponds to 48.6%. Therefore, the percentage dissociation is:

\[ \text{Percentage dissociation} = 48.6\% \times 100 = 90\% \] Thus, the correct answer is option (A), 90%.