
Calculate Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] having a degree of dissociation $\left( \alpha \right)$ equal to \[0.778.\]
A. \[4.334\]
B. \[3.334\]
C. \[0.222\]
D. \[2.334\]
Answer
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Hint: The degree of dissociation is the method of formation of free ions which carry the current which is represented as $\left( \alpha \right)$. And these free ions are dissociated from the fraction of solute at the presence of a given concentration. The fraction of molecules present in the substance will dissociate at a particular time. Potassium ferricyanide is a chemical element having the formula \[{K_3}[Fe{(CN)_6}]\].
Complete answer:
The Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is not equal to \[4.334\]. Hence, option (A) is incorrect.
According to the question, the degree of dissociation of aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is equal to \[0.778.\]
The formula used to find out the Van’t Hoff factor is,
\[i = 1 + (n - 1)\alpha \] --- (1)
Where, i is equal to Van’t Hoff factor
n is equal to number of ions forms from the substance and
$\alpha $ is equal to degree of dissociation.
Here the given substance is potassium ferricyanide and it is dissociated into, \[{K^ + }\] ions and \[{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}\]. Let’s see the equation,
\[{K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \to 3{K^ + } + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}\]
Here, there is a formation of four ions by the dissociation of potassium ferricyanide. So, n is equal to four. Substitute these values in equation one will get,
\[i = 1 + (4 - 1) \times 0.778\]
On simplification we get,
\[i = 3.334\]
Therefore, Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is equal to \[3.334\]. Hence, option (B) is correct.
The Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is not equal to \[0.222\]. Hence, option (C) is incorrect.
The Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is not equal to \[2.334\]. Hence, option (D) is incorrect.
Hence, option (B) is correct.
Note:
We need to know that the Van’t Hoff factor is a chemical term which is represented by ‘i’ and it can be found out by using degree of dissociation and number of ions formed from the substance. And this factor calculates how solute affects the colligative properties like freezing-point depression, elevation of boiling point, osmotic pressure and vapour pressure. Hence, this factor can be applicable for any colligative properties.
Complete answer:
The Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is not equal to \[4.334\]. Hence, option (A) is incorrect.
According to the question, the degree of dissociation of aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is equal to \[0.778.\]
The formula used to find out the Van’t Hoff factor is,
\[i = 1 + (n - 1)\alpha \] --- (1)
Where, i is equal to Van’t Hoff factor
n is equal to number of ions forms from the substance and
$\alpha $ is equal to degree of dissociation.
Here the given substance is potassium ferricyanide and it is dissociated into, \[{K^ + }\] ions and \[{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}\]. Let’s see the equation,
\[{K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \to 3{K^ + } + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}\]
Here, there is a formation of four ions by the dissociation of potassium ferricyanide. So, n is equal to four. Substitute these values in equation one will get,
\[i = 1 + (4 - 1) \times 0.778\]
On simplification we get,
\[i = 3.334\]
Therefore, Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is equal to \[3.334\]. Hence, option (B) is correct.
The Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is not equal to \[0.222\]. Hence, option (C) is incorrect.
The Van't Hoff factor (i) for an aqueous solution of \[{K_3}[Fe{(CN)_6}]\] is not equal to \[2.334\]. Hence, option (D) is incorrect.
Hence, option (B) is correct.
Note:
We need to know that the Van’t Hoff factor is a chemical term which is represented by ‘i’ and it can be found out by using degree of dissociation and number of ions formed from the substance. And this factor calculates how solute affects the colligative properties like freezing-point depression, elevation of boiling point, osmotic pressure and vapour pressure. Hence, this factor can be applicable for any colligative properties.
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