JEE MainChemistryModeratePhysical Chemistry

The freezing point (in ºC) of a solution containing 0.1 g of
K3[Fe(CN)6] (Mol. Wt. 329) in 100 g of water (Kf = 1.86 K kg mol-1)
is:
100 g ty (Kf = 1.86 K kg mol-1) esa 0.1 g K3[Fe(CN)6] (v.kq Hkkj =
329) ds ?kqyus

Show answer and solution

\Delta T_{f} = i \times K_{f} \times m

= 4 \times 1.86 \times \frac{0.1}{329 \times 0.1} = 2.3 \times 10^{- 2}

\Rightarrow T_{f} = 0 - 2.3 \times 10^{- 2} = - 2.3 \times 10^{- 2}\ ^{\circ}C

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx

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