The observed osmotic pressure for a 0.10 M s | vedclass

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(A) The dissociation of $Fe(NH_4)_2(SO_4)_2$ is given by: $Fe(NH_4)_2(SO_4)_2 \rightarrow Fe^{2+} + 2NH_4^+ + 2SO_4^{2-}$.The total number of ions pro

Vedclass · Accepted Answer

$5$ and $4.42$

Vedclass · Answer

(A) The dissociation of $Fe(NH_4)_2(SO_4)_2$ is given by: $Fe(NH_4)_2(SO_4)_2 ightarrow Fe^{2+} + 2NH_4^+ + 2SO_4^{2-}$.The total number of ions produced per formula unit is $1 + 2 + 2 = 5$. Thus,the expected van't Hoff factor $(i_{expected}) = 5$.The normal osmotic pressure $(\pi_{nor})$ is calculated using the formula $\pi = CRT$:$\pi_{nor} = 0.10 	imes 0.082 	imes 298 = 2.4436 \ atm$.The observed osmotic pressure $(\pi_{ob})$ is given as $10.8 \ atm$.The experimental van't Hoff factor $(i_{exp})$ is calculated as:$i_{exp} = \frac{\pi_{ob}}{\pi_{nor}} = \frac{10.8}{2.4436} \approx 4.42$.Therefore,the expected and experimental values are $5$ and $4.42$ respectively.

The observed osmotic pressure for a $0.10 \ M$ solution of $Fe(NH_4)_2(SO_4)_2$ at $25 \ ^oC$ is $10.8 \ atm$. The expected and experimental (observed) values of van't Hoff factor $(i)$ will be respectively: $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

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