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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Chemistry
Chapter
Solutions

JEE Advanced 2011 · Paper 2 · Chemistry · Solutions

The freezing point (in ^ C ) of solution containing 0.1 ~g of...

The freezing point (in ∘C{ }^{\circ} \mathrm{C} ) of solution containing 0.1 g0.1 \mathrm{~g} of K3[Fe(CN)6]\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] (molecular weight 329)) in 100 g100 \mathrm{~g} of water \left(K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$ is

Options

  1. A−2.3×10−2-2.3 \times 10^{-2}
  2. B−5.7×10−2-5.7 \times 10^{-2}
  3. C−5.7×10−3-5.7 \times 10^{-3}
  4. D−1.2×10−2-1.2 \times 10^{-2}

Correct answer

A. −2.3×10−2-2.3 \times 10^{-2}

Step-by-step solution

vant Hoff's factor (i)=4{3 K++[Fe(CN)6]3−}(i)=4\left\{3 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\right\} \quad Molality =0.1329×1000100=1329=\frac{0.1}{329} \times \frac{1000}{100}=\frac{1}{329} \Rightarrow \quad-Δ\DeltaT_f=i K_f ⋅\cdotm=4 ×\times1.86 ×1329=\times \frac{1}{329}=2.3 ×\times10^{-2} \Rightarrow T_f=-2.3 ×\times10^{-2}{ }^{{}^{\circ}} C(\mathrm{C} (As freezing point of water is 0∘C0^{{}^{\circ}} \mathrm{C} )
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