Question

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

• A. $5$ and $4.42$
• B. $4$ and $4.00$
• C. $5$ and $3.42$
• D. $3$ and $5.42$

Answer 1

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

AS we know,$Fe\left(N{H}_4{)}_2\right(S{O}_4{)}_2\to 2N{H}_4^{+}+F{e}^{+}+2S{O}_4^{2-}$
Expected value if $i=5$
Now, observed van't hoff factor $=\frac{P}{MRT}$
$=\frac{10.8}{0.10\times 0.082\times 298}$
$=4.42$
Hence, option (a) is correct.