Question

A decimolar solution of potassium ferrocynide is $50\%$ dissociated at 300 K. Given $S=8.314J{K}^{-1}mo{l}^{-1}$. The osmotic pressure of the solution is :

• A. $7.438\times 105N{m}^{-2}$
• B. $7.834\times 105N{m}^{-2}$
• C. $7.348\times 105N{m}^{-2}$
• D. $7.483\times 105N{m}^{-2}$

Answer 1

The correct option is D $7.483\times 105N{m}^{-2}$

We have $molarity=\frac{n}{V}=0.1mollitr{e}^{-1}=\frac{0.1}{{10}^{-3}}mol{m}^{-3}={10}^{2}mol{m}^{-3}As,\pi N\times V=nSTOr,{\pi }_N=\left(\frac{n}{V}\right)ST={10}^2\times 8.314\times 300N{m}^{-2}Letdegreeofassociationbe\alpha ,Therefore,for{K}_{4}Fe(CN{)}_6\leftrightarrow 4K^{+}+Fe(CN{)}_6^{4-}$
1 0 0
$1-\alpha$ $4\alpha$ $\alpha$
It is given that $\alpha =0.5$
And $\frac{{\pi }_{exp}}{{\pi }_N}=1+4\alpha Therefore,{\pi }_{exp}={\pi }_N(1+4\alpha )=={10}^2\times 8.314\times 300\times (1+4\times 0.5)={10}^2\times 8.314\times 300\times (1+2)=7.483\times 105N{m}^{-2}$