Question

What is the value of i for k4[fe(cn)6] which is 80% dissociated?

Answer 1

K4[Fe(CN)6]•3H2O => 4K+ + [Fe(CN)6]4-. 1 0 0 (initial). 1-x 4x x (atfter dissociation). vant hoff factor .

Answer 2

Answer:

$4.2$is the value of i for k4[fe(cn)6] which is 80% dissociated.

Explanation:

$\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$

Since one molecule of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$dissociates to produce 5 ions, the value of van't Hofff factor is $5 .$

$\alpha=\frac{i-1}{n-1}$

$\mathrm{n}=5 , since \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] gives 5 ions in the solution.$

$\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightarrow 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$

Now,

$\therefore 0.8=\frac{\mathrm{i}-1}{5-1}$

$\mathrm{i}=4.2$