A solution of \( 1.25 \mathrm{~g} \) of ' \( \mathrm{P} \) ' in \( 50 \mathrm{~g} \) of water lowers freezing point by \( 0.3{ }^{\circ} \mathrm{C} \). Molar mass of ' \( \mathrm{P} \) ' is
94. \( K_{\mathrm{f}(\text { water })}=1.86 \mathrm{Kg} \mathrm{mol}^{-1} \). The degree of association of ' \( \mathrm{P} \) ' in water is

Given,
\(w_{B}=1.25 \mathrm{~g}\)
\(w_{A}=\) mass of solvent \(=50 \mathrm{~g}\)
\(T_{f}=0.3^{\circ} \mathrm{C}\)
Molecular mass of \(\mathrm{P}=94\)
\(K_{f}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
\(\Delta T_{f}=0-(-0.3)=0.3^{\circ} \mathrm{C}\)
\(M_{B}=\frac{K_{f} \times w_{B} \times 1000}{\Delta T_{f} \times w_{A}}\)
\(M_{B}=\frac{1.86 \times 1.25 \times 1000}{0.3 \times 50}=155\)
Now, van't Hoff factor,
\(i=\frac{\text { Normal molar mass }}{\text { Observed molar mass }}\)
\(=\frac{94}{155}=0.6064\)
Initial moles
Moles after association \(1-\alpha \mathrm{a} / 2\)
(If \(a\) is the degree of association)
Degree of association \((\alpha)=\frac{n(1-i)}{n-1}=\frac{2 \times(1-0.6064)}{1}\)
\(=78.7 \%\)