What will be the osmotic pressure of 0.2 M K | vedclass

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(A) The formula for osmotic pressure is $\pi = iCRT$. For $K_4[Fe(CN)_6]$,the van't Hoff factor $i = 5$ (since it dissociates as $4K^+ + [Fe(CN)_6]^{4

Vedclass · Accepted Answer

$24.6$

Vedclass · Answer

(A) The formula for osmotic pressure is $\pi = iCRT$. For $K_4[Fe(CN)_6]$,the van't Hoff factor $i = 5$ (since it dissociates as $4K^+ + [Fe(CN)_6]^{4-}$). Given concentration $C = 0.2 \ M$,gas constant $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and temperature $T = 27 + 273 = 300 \ K$. Substituting the values: $\pi = 5 	imes 0.2 	imes 0.0821 	imes 300 = 1 	imes 0.0821 	imes 300 = 24.63 \ atm$. Thus,the osmotic pressure is approximately $24.6 \ atm$.

What will be the osmotic pressure of $0.2 \ M$ $K_4[Fe(CN)_6]$ at $27^{\circ} C$ (in $atm$)?

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