Q54 / 180 · KCET 2014 · Chemistry
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
94. \( K_{\mathrm{f}(\text { water })}=1.86 \mathrm{Kg} \mathrm{mol}^{-1} \). The degree of association of ' \( \mathrm{P} \) ' in water is
- A\( 60 \% \)
- B\( 75 \% \)
- C\( 80 \% \)
- Correct answerDNone of the above
Solution
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 $(\alpha)=\frac{n(1-i)}{n-1}=\frac{2 \times(1-0.6064)}{1}$
$=78.7 \%$
Hence, the correct answer is option: 'D'.