Question

A decimolar solution of potassium ferrocyanide is 50% dissociated at 300K. Calculate the osmotic pressure of the solution. $(R=8.314J{K}^{-1}mo{l}^{-1})$

• A. $3.74\times {10}^{5}N{m}^{-2}$
• B. $7.482\times {10}^{5}N{m}^{-2}$
• C. $14.96\times {10}^{5}N{m}^{-2}$
• D. $14.96\times {10}^{6}N{m}^{-2}$

Answer 1

Answer: (B)

$7.482\times {10}^{5}N{m}^{-2}$

$K_4\left[Fe\right(CN{)}_6]\rightleftharpoons 4K^{+}+[Fe(CN{)}_6{]}^{4-}$
Since the solution is $50\%$ dissociated, the Vant Hoff's factor is as follows:

$i=[1+(n^{\prime }-1\left)\alpha \right]=[1+(5-1\left)0.5\right]=1+2=3$
The osmotic pressure, $\Pi =iCRT$

Here, $i$ is Vant Hoff's factor, $\alpha$ is the degree of dissociation, $C$ is the molar concentration $\left(0.1\right)$, $R$ is the ideal gas constant $(0.08206$ L atm / mol. K$)$, $T$ is the absolute temperature, $300K$

$\Pi =3\times 0.1\times 0.08206\times 300=7.385atm=7.482\times {10}^{5}N{m}^{-2}$
$\because 1atm=101325N{m}^{-2}$