Question

0.1M K4[Fe(CN)6] is 60% ionized.What will be the Vant Hoffs Factor​

Answer 1

Answer:

i = 25

Explanation:

$............................................K_{4}[Fe(CN)_{6}] --> 4K^{+} + [Fe(CN)_{6}]^{4-}\\Before Dissociation --> .........0.1.....................0..................0 \\After Dissociation --> .....0.1-\alpha .................4\alpha .................\alpha$

Van't Hoff Factor (i) = number of moles after dissociation/ number of moles before dissociation

i = $\frac{ 4\alpha + \alpha + 0.1 -\alpha}{0.1 +0+0} = \frac{ 4\alpha + 0.1}{0.1}$.....where $\alpha =60/100 = 0/6$

Upon solving we get i=25

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