JEE Advanced · Chemistry · 16. Solutions
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
- A \(-2.3 \times 10^{-2}\)
- B \(-5.7 \times 10^{-2}\)
- C \(-5.7 \times 10^{-3}\)
- D \(-1.2 \times 10^{-2}\)
Answer & Solution
Correct Answer
(A) \(-2.3 \times 10^{-2}\)
Step-by-step Solution
Detailed explanation
vant Hoff's factor \((i)=4\left\{3 \mathrm{~K}^{+}+\left(\mathrm{Fe}(\mathrm{CN})_6\right)^{3-}\right\} \quad\) Molality \(=\frac{0.1}{329} \times \frac{1000}{100}=\frac{1}{329}\)
\(\Rightarrow -\Delta T_f=i K_f \cdot m=4 \times 1.86 \times \frac{1}{329}\) \(=2.3 \times 10^{-2} \Rightarrow T_f=-2.3 \times 10^{-2}{ }^{\circ} \mathrm{C}\)
(As freezing point of water is \(0^{\circ} \mathrm{C}\) )
\(\Rightarrow -\Delta T_f=i K_f \cdot m=4 \times 1.86 \times \frac{1}{329}\) \(=2.3 \times 10^{-2} \Rightarrow T_f=-2.3 \times 10^{-2}{ }^{\circ} \mathrm{C}\)
(As freezing point of water is \(0^{\circ} \mathrm{C}\) )
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