A decimolar solution of $K_{4} [F e (C N)_{6}]$ is $50 %$ dissociated at $300 K$ . Calculate the Osmotic pressure of solution.

Open in App

Solution

Degree of dissociation, $x = 50 % = 0.5$
Van't Hoff factor, $i = \frac{n u m b e r o f m o l e s a f t e r d i s s o c i a t i o n}{t o t a l n u m b e r o f m o l e s i n i t i a l l y} = \frac{1 + 4 x}{1}$
$i = 1 + 4 (0.5) = 1 + 2 = 3$
Decimolar solution means the molarity of solution is M/10 or 0.1 M. That is, 0.1 mole of solute is dissolved in 1 L of solution. Since, $1 L = 1 / 1000 m^{3}$ , therefore it can be said, 0.1 mole of solute is dissolved in $1 / 1000 m^{3}$ volume of solution.
We have, $π V = i n R T$
$π = \frac{n}{V} R T$
$π = o s m o t i c p r e s s u r e$
$V = v o l u m e o f s o l u t i o n$
$n = n u m b e r o f m o l e s o f s o l u t e$
$R = u n i v e r s a l g a s c o n s t a n t = 8.314 N m m o l^{- 1} K^{- 1}$
$T = t e m p e r a t u r e = 300 K$
$i = V a n^{'} t H o f f f a c t o r = 3$
$π = 3 \times \frac{0.1}{0.001} \times 8.314 \times 300$
$π = 7.48 \times 10^{5} N m^{- 2} = \frac{7.48 \times 10^{5}}{1.013 \times 10^{5}} a t m$
$π = 7.38 a t m$

Suggest Corrections

Similar questions

K_{4} [F e {(C N)}_{6}]

50

300 K

K_{4} [F e {(C N)}_{6}]

50 %

300

(R = 8.314 J K^{- 1} m o l^{- 1})

0.1 M

K_{4} [F e (C N)_{6}]

50 %

27^{\circ} C

300 K

50 %

R = 0.082 L a t m m o l^{- 1} K^{- 1}

Join BYJU'S Learning Program