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The freezing point depression of .001 …

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The freezing point depression of .001 M of Kx[Fe(CN)6] is 7.10 * 10-3K. Determine the value of x. Kf=1.86 K kg /mole for water.

    • Posted by Sahil Shakoor 5 years, 7 months ago

    ICSE & ISC > Class 12 > Chemistry
    • 1 answers

    Gaurav Seth 5 years, 7 months ago

     

    Freezing point depressionΔTf = i × Kf × mi = ΔTf Kf × m    = 7.10 × 10−31.86 × 0.001    =3.82 ≈ 4The total number of value of i = 4The given complex Kx[Fe(CN)6]  has already complex ion [Fe(CN)6] as 1.So the remaining are K ions = 4−1 = 3Thus  the value of x =3 and the complex is K3[Fe(CN)6]   

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