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(A) The van't Hoff factor $(i)$ is defined as the ratio of the theoretical (normal) molecular mass to the observed (experimental) molecular mass: $i =

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$1$

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(A) The van't Hoff factor $(i)$ is defined as the ratio of the theoretical (normal) molecular mass to the observed (experimental) molecular mass: $i = \frac{	ext{Molecular mass (theoretical)}}{	ext{Molecular mass (experimental)}}$ Given that the experimental molecular mass is double the actual (theoretical) value,we have: $i = \frac{1}{2} = 0.5$ For the association of a solute to form a dimer,the relationship between the van't Hoff factor $(i)$ and the degree of association $(\alpha)$ is given by: $i = 1 - \alpha + \frac{\alpha}{n}$ Since it is a dimer,$n = 2$: $0.5 = 1 - \alpha + \frac{\alpha}{2}$ $0.5 = 1 - \frac{\alpha}{2}$ $\frac{\alpha}{2} = 1 - 0.5 = 0.5$ $\alpha = 1$ Thus,the degree of association is $1$ (or $100\%$).

In an experiment to estimate the molecular weight of benzoic acid by the elevation in boiling point method,the experimental value of the molecular weight was double the actual value. Calculate the degree of association of the dimer.

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