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

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]

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

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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The freezing point depression of 0.001 m $K_{x} [Fe {(CN)}_{6}]$ is $7.10 \times 10^{- 3} K$ . Determine the value of x. Given, $K_{f} = 1 .86 K kg {mol}^{- 1}$ for water

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$\begin{array}{l} Δx = i \times K_{f} \times m \\ 7 .10 \times 10^{- 3} = i \times 1 .86 \times 0 .001 \\ i = 3 .817 \\ α = \frac{i - 1}{n - 1} \\ 1 = \frac{3 .817 - 1}{(x + 1) - 1} \\ x = 2 .817 = 3 \end{array}$

$∴$ Molecular formula of the compound is

$K_{3} [Fe {(CN)}_{6}]$

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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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The freezing point depression of 0.001 m $K_{x} [Fe {(CN)}_{6}]$ is $7.10 \times 10^{- 3} K$ . Determine the value of x. Given, $K_{f} = 1 .86 K kg {mol}^{- 1}$ for water