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Answer the following.
When 1·90 g of benzoic acid is dissolved in 26 g of benzene, it causes a depression in freezing…
When $1\cdot90\text{ g}$ of benzoic acid is dissolved in $26\text{ g}$ of benzene, it causes a depression in freezing point by $1\cdot60\text{ K}$. The molal depression constant for benzene is $4\cdot8\text{ K kg mol}^{-1}$ and the molecular mass of benzoic acid is $122\text{ g mol}^{-1}$.
Calculate the degree of association of benzoic acid if it forms dimer in solution.
Answer
Answer
$m=\frac{1.90/122}{26/1000}=0.599\text{ mol kg}^{-1}$; $i=\frac{\Delta T_f}{K_f m}=\frac{1.60}{4.8\times0.599}=0.5565$. For dimerisation $i=1-\frac{\alpha}{2}$, so $\alpha=2(1-0.5565)=0.887$ (88.7%)
Final answer: 0.887