Explanation:
(a) Given:\(\mathrm{K}_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, \mathrm{w}=1.25 \mathrm{~g}\),
\(\mathrm{W}=50 \mathrm{~g}, \Delta \mathrm{~T}=0.3^{\circ} \mathrm{C}, \mathrm{M}=\) ?
As \(P\) undergoes association,
\(2 \mathrm{P} \rightleftharpoons(\mathrm{P})_2 ; \mathrm{n}=2\)
Using equation, \(\mathrm{M}=\frac{1000 \times \mathrm{K}_{\mathrm{f}} \times \mathrm{w}}{\mathrm{W} \times \Delta \mathrm{T}}\)
\(\mathrm{M}=\frac{1000 \times 1.86 \times 1.25}{50 \times 0.3}=155\)
Now, \(\mathrm{i}=\frac{\text { Normal mol. mass }}{\text { Observed mol. mass }}=\frac{94}{155}=0.606\)
Then, the degree of association of \(P\) is
\(\alpha=\frac{1-\mathrm{i}}{1-\frac{1}{n}}\)
\(\alpha=\frac{1-0.606}{1-\frac{1}{2}}=0.788\) or \(78.8 \% \sim 80 \%\)