Question

M/10 Solution Of Potassium Ferrocyanide Is 6% Dissociated At .What Will Be Its Osmotic Pressure?

Answer 1

Answer:

188.29 Pa

Explanation:

Refer attachment for explanation!!!!

Answer 2

Answer:

The Osmotic pressure of potassium Ferrocyanide is $\bold{7.38 atm}$ when it is dissociated by $6\%$ .

Explanation:

The dissociation reaction of potassium ferrocyanide is

$K_4[Fe(CN)_6]\rightarrow4K^++[Fe(CN)_6]^4^-$

So, vn Hoff's factor is

$i=1-\alpha+4\alpha+\alpha\\ =1+4\alpha$

But $\alpha$ is given $50\%$ . so,  $\alpha=0.5$

Now,

$i=1+4\alpha\\=1+4\times0.5\\=1+2\\=3$

Given that C $=0.1M$

                $T=300K$

                R $=0.082 atm.L/K/mol$

Now, Osmotic pressure is

$\pi =iCRT\\=3\times0.1\times0.082\times300\\=900\times0.0082\\=7.38 atm$  

Hence, the Osmotic pressure of potassium Ferrocyanide is $\bold{7.38 atm}$ when it is dissociated by $6\%$ .