Question

A $0.1M$ solution of $K_4\left[Fe\right(CN{)}_6]$ is $50\%$ dissociated at ${27}^{\circ }C$. What will be its observed osmotic pressure?

• A. $5.1atm$
• B. $2.9atm$
• C. $7.4atm$
• D. $9.1atm$

Answer 1

The correct option is C $7.4atm$

$i=\frac{\text{Experimental Colligative Property (observed)}}{\text{Calculated Colligative Property (expected)}}$

$\text{Observed osmotic pressure}:{\pi }_{obs}=i\times C\times R\times T$
Here $C=\text{Concentration}=0.1MR=0.0821Latmmo{l}^{-1}{K}^{-1}T={27}^{\circ }C=300K$

Degree of dissociation $(\alpha =0.5)$
$i=1+(n-1)\alpha$

For $K_4\left[Fe\right(CN{)}_6]$ :

$K_4\left[Fe\right(CN{)}_6]\to 4K^{+}+[Fe(CN{)}_6{]}^{4-}n=5$
$i=1+(5-1)\times 0.5i=3$
So,
${\pi }_{obs}=(3\times 0.1\times \times 0.0821\times 300)atm{\pi }_{obs}=7.389atm\approx 7.4atm$