Kerala KEAM 5 June 2024 Solved Paper

The observed molecular weight $M_{obs}$ of an electrolyte can be related to the theoretical molecular weight $M_{theo}$ and the degree of dissociation $\alpha$ using the formula: $$M_{obs} = \frac{M_{theo}}{1 + \alpha (n - 1)}$$where:- $M_{obs}$ is the observed molecular weight,- $M_{theo}$ 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:- $M_{obs} = 117 \, g/mol$,- $M_{theo} = 60 \, g/mol$,- $n = 2$.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: $$Percentage dissociation = 48.6\% \times 100 = 90\%$$Thus, the correct answer is option (A), 90%. Quick Tip: To calculate the percentage dissociation of an electrolyte, use the formula relating observed and theoretical molecular weights, and solve for the degree of dissociation.