Question

The observed osmotic pressure for a $0.10\text{M}$ solution of $Fe\left(N{H}_4\right)2(S{O}_4{)}_2$​ at $25°C$ is $10.8\text{atm}$. The expected and experimental (observed) values of Van't Hoff factor (i) will be respectively:
$[R=0.082L{\text{atm K}}^{-1}{\text{mol}}^{-1}]$

• A. $5$ and $4.42$
• B. $4$ and $8.42$
• C. $5$ and $3.42$
• D. $4$ and $5.42$

Answer 1

The correct option is A $5$ and $4.42$

We know, in solution $Fe\left(N{H}_4\right)2(S{O}_4{)}_2$ will dissociates as: $Fe\left(N{H}_4{)}_2\right(S{O}_4{)}_2\to 2N{H}_4^{+}+F{e}^{2+}+2S{O}_4^{2-}$
So, expected value of $i=5$
Again, $\text{Osmotic pressure}\Pi =iCRT$
$\text{Osmotic pressure}\Pi =10.8\text{atm}$
$\text{Temperature},T=298\text{K}$
We are given, $\text{Concentration},C=0.10\text{M}$
Therefore, observed van't Hoff factor $i=\frac{\Pi }{CRT}$

$=\frac{10.8}{0.10\times 0.082\times 298}\approx 4.42$
Hence, option (A) is correct.