The freezing point depression of 0.001 m Kx[Fe(CN)6] is 7.10×10−3 K. Determine the value of x. Given, Kf=1.86 K kg mol−1 for water

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answer is A.

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Detailed Solution

Δx=i×Kf×m7.10×10−3=i×1.86×0.001i=3.817α=i−1n−11=3.817−1(x+1)−1x=2.817=3∴ Molecular formula of the compound is K3[Fe(CN)6]

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