Question

A 0.004M solution of K4[Fe(CN)6] is isotonic with 0.01M solution of urea at same temperature. The molarity of K+ in the solution will be???
1)0.375M
2)3M
3)0.006M
4)0.015M

Correct answer is option 3
pls provide the solution for it???

Answer 1

Answer : The correct option is, (3) 0.006 M

Explanation :

Isotonic solutions are those solutions which have the same osmotic pressure.

$\pi =iCRT$ if osmotic pressures are equal at the same temperature, concentrations must also be equal.

where,

i = Van't Hoff factor

For non-electrolyte i = 1 (always) and urea is a non-electrolyte.

$\pi$ = osmotic pressure

C = concentration

R = solution constant

T = temperature

Thus,

$C_{urea}=i\times C_A$ ...........(1)

where,

$C_{urea}$ = concentration or molarity of urea = 0.01 M

$C_A$ = concentration or molarity of $K_4[Fe(CN)_6]$ = 0.004 M

Now put all the given values in the above expression 1, we get :

$0.01M=i\times 0.004M$

$i=2.5$

Now we have to calculate the degree of dissociation.

The complete dissociation reaction will be,

               $K_4[Fe(CN)_6]\rightleftharpoons 4K^++[Fe(CN)_6]^{-1}$

initially            c                     0              0

At eqm.        $c(1-\alpha)$           $4c\alpha$          $c\alpha$

$\alpha=\frac{i-1}{n-1}$

where,

$\alpha$ = degree of dissociation

n = number of ions on complete dissociation = 5

Now put all the given values in the above formula, we get :

$\alpha=\frac{2.5-1}{5-1}=0.375$

Now we have to calculate the molarity of $K^+$ ion in the solution.

From the balanced reaction, we conclude that 1 mole of $K_4[Fe(CN)_6]$ complex on complete dissociation gives 4 moles of $K^+$ ion.

Molarity of $K^+$ ion = $4c\alpha$ = $4\times 0.004M\times 0.375=0.006M$

Therefore, the molarity of $K^+$ ion in the solution will be, 0.006 M