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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
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The correct answer is:
$-2.3 \times 10^{-2}$
$-2.3 \times 10^{-2}$
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 $0^{\circ} \mathrm{C}$ )
$$
\Rightarrow \quad-\Delta T_f=i K_f \cdot m=4 \times 1.86 \times \frac{1}{329}=2.3 \times 10^{-2} \Rightarrow T_f=-2.3 \times 10^{-2}{ }^{\circ} \mathrm{C}
$$
(As freezing point of water is $0^{\circ} \mathrm{C}$ )
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