Question:
At \(300\ \mathrm{K}\), a decimolar solution of potassium ferrocyanide is \(50\%\) dissociated. The osmotic pressure (in atm) of the solution is
\[
(R=0.082\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}})
\]
At \(300\ \mathrm{K}\), a decimolar solution of potassium ferrocyanide is \(50\%\) dissociated. The osmotic pressure (in atm) of the solution is
\[
(R=0.082\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}})
\]
Show Hint
For electrolytes,
\[
\boxed{i=1+\alpha(n-1)}
\]
where
\begin{itemize}
\item \(\alpha\) = degree of dissociation,
\item \(n\) = total number of ions produced.
\end{itemize}
Then use
\[
\boxed{\pi=iCRT.}
\]
Updated On: Jul 9, 2026
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- 7.38
- 7.83 \bigskip
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
Osmotic pressure is given by
\[
\boxed{\pi=iCRT}
\]
where \(i\) is the van't Hoff factor.
Step 1: Calculate the van't Hoff factor. Potassium ferrocyanide dissociates as \[ \mathrm{K_4[Fe(CN)_6]} \rightarrow 4\mathrm{K^+} +\mathrm{[Fe(CN)_6]^{4-}} \] Thus, \[ n=5 \] Since degree of dissociation, \[ \alpha=0.5 \] \[ i=1+\alpha(n-1) \] \[ =1+0.5(5-1) =3 \]
Step 2: Calculate osmotic pressure. Given, \[ C=0.1\ \mathrm{M},\qquad R=0.082,\qquad T=300\ \mathrm{K} \] \[ \pi =3\times0.1\times0.082\times300 =7.38\ \mathrm{atm} \]
Step 3: Final conclusion. \[ \boxed{7.38\ \mathrm{atm}} \] Hence, the correct option is \(\boxed{(C)}\).
Step 1: Calculate the van't Hoff factor. Potassium ferrocyanide dissociates as \[ \mathrm{K_4[Fe(CN)_6]} \rightarrow 4\mathrm{K^+} +\mathrm{[Fe(CN)_6]^{4-}} \] Thus, \[ n=5 \] Since degree of dissociation, \[ \alpha=0.5 \] \[ i=1+\alpha(n-1) \] \[ =1+0.5(5-1) =3 \]
Step 2: Calculate osmotic pressure. Given, \[ C=0.1\ \mathrm{M},\qquad R=0.082,\qquad T=300\ \mathrm{K} \] \[ \pi =3\times0.1\times0.082\times300 =7.38\ \mathrm{atm} \]
Step 3: Final conclusion. \[ \boxed{7.38\ \mathrm{atm}} \] Hence, the correct option is \(\boxed{(C)}\).
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