Question

what is vant hoff factor in k4[Fe(CN)6]and bacl2

Answer 1

The Van’t Hoff Factor is essentially a fraction or ratio of the number of dissociated ions when a solute is dissolved in a medium to the number of molecules which gave rise to that number of dissociated ions.

By this theory, the van’t Hoff Factor of what is $\bold{\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \text { and } \mathrm{BaCl}_{2}}$ are as follows:

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

Here, Van’t Hoff Factor is (4+1) ∶ 1 = 5

$\mathrm{BaCl}_{2} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}$

Here, van’t Hoff Factor is (1+2) ∶ 1 = 3

Answer 2

Answer:

i for k4 fe(Cn)6 = 5

Explanation:

i for k4 fe(Cn)6 = (4 of K) + (1 of fe(cn)6)= 5

i for bacl2 = 1 of ba + 2 of cl = 3