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In an experiment to es timate the molecular weight of benzoic acid by elevation in boiling point method, the experimental value of molecular weight was double the actual value. Calculate the degree of association of dimer, if the elevation in B.P is $2^{\circ} \mathrm{C}$.
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$\because$ According to van't-Hoff volume $(i)$
$$
(i)=\frac{\text { Molecular mass (calculated) }}{\text { Molecular mass (experimental) }}
$$
Given, molecular mass (experimental) is double.
$$
\therefore \quad i=\frac{1}{2}
$$
Also,
$$
\because \quad(i)=(1-\alpha)+\frac{\alpha}{n}
$$
where, $\alpha=$ degree of association and $n=$ number of particles associated.
(Here, $n=2, \because i=1 / 2$ )
Therefore,
$\frac{1}{2}=(1-\alpha)+\frac{\alpha}{2}$
$$
\frac{1}{2}=1-\frac{\alpha}{2}
$$
or, $\frac{\alpha}{2}=\frac{1}{2}$
or, $\alpha=1$
$$
(i)=\frac{\text { Molecular mass (calculated) }}{\text { Molecular mass (experimental) }}
$$
Given, molecular mass (experimental) is double.
$$
\therefore \quad i=\frac{1}{2}
$$
Also,
$$
\because \quad(i)=(1-\alpha)+\frac{\alpha}{n}
$$
where, $\alpha=$ degree of association and $n=$ number of particles associated.
(Here, $n=2, \because i=1 / 2$ )
Therefore,
$\frac{1}{2}=(1-\alpha)+\frac{\alpha}{2}$
$$
\frac{1}{2}=1-\frac{\alpha}{2}
$$
or, $\frac{\alpha}{2}=\frac{1}{2}$
or, $\alpha=1$
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