Q5 / 60 · JEE Advanced 2011 (Paper 2) · Chemistry
The freezing point (in ${ }^{\circ} \mathrm{C}$ ) of solution containing $0.1 \mathrm{~g}$ of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ (molecular weight 329$)$ in $100 \mathrm{~g}$ of water $\left(K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$ is
- Correct answerA$-2.3 \times 10^{-2}$
- B$-5.7 \times 10^{-2}$
- C$-5.7 \times 10^{-3}$
- D$-1.2 \times 10^{-2}$
Solution
vant Hoff's factor $(i)=4\left\{3 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\right\} \quad$ Molality $=\frac{0.1}{329} \times \frac{1000}{100}=\frac{1}{329}$
$$
\Rightarrow \quad-\Delta T_f=i K_f \cdot m=4 \times 1.86 \times \frac{1}{329}=2.3 \times 10^{-2} \Rightarrow T_f=-2.3 \times 10^{-2}{ }^{\circ} \mathrm{C}
$$
(As freezing point of water is $0^{\circ} \mathrm{C}$ )
Hence, the correct answer is option: 'A'.