\[{K_2}Hg{I_4}\]is \[40\% \] ionised in aqueous solution .The value of its van’t Hoff factor \[(i)\] is:
(A) \[1.8\]
(B) \[2.2\]
(C) \[2.0\]
(D) \[1.6\]
Answer
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Hint: As we know that \[{K_2}Hg{I_4}\]is the ionic solute and dissociated into solvent into its ion which obeys the law of conservation of mass. The value of van’t Hoff totally depends upon the dissociation of \[{K_2}Hg{I_4}\]. The van’t-Hoff factor is used to modify the calculation of colligative property.
Complete step by step answer:
When any compound dissociates into its ions then we use a degree of dissociation which is represented by \[\alpha \].
When \[{K_2}Hg{I_4}\]is dissolved in solvent it gives ions as
\[{K_2}Hg{I_4} \to 2{K^ + }(aq) + Hg{I_4}^{2 - }(aq)\]
The number of ions \[\left( n \right) = 3\]
And the degree of dissociation (\[\alpha \]) is given as\[ = 0.40\]
So, we can write an equation as
\[\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_2}Hg{I_4} \to 2{K^ + }(aq) + Hg{I_4}^{2 - }(aq)\\
initial\,moles\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\
moles\,after\,dissociation\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha
\end{array}\]
So, the total number of moles of ions and the undissociated \[{K_2}Hg{I_4}\]molecules\[ = 1 - \alpha + 2\alpha + \alpha = 2\alpha + 1\]
The its van’t Hoff factor \[(i)\]
\[i = \dfrac{{total\,number\,of\,moles\,}}{{initial\,moles}}\]
By Putting the values of total number of moles we get,
\[\begin{array}{l}
i = \dfrac{{2\alpha + 1}}{1}\\
i = \dfrac{{0.80 + 1}}{1}\\
i = 1.80\\
i > 1
\end{array}\]
So, by using van’t Hoff factor \[(i)\],we can also calculate the extent of dissociation or association.
Therefore, the correct option is option (A).
Note:
The Van't Hoff factor is also defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property. Further, as the colligative property is inversely proportional to the molecular mass of the solute, we can also write that the van't Hoff factor is the ratio of normal molecular mass to the abnormal molecular mass.
Complete step by step answer:
When any compound dissociates into its ions then we use a degree of dissociation which is represented by \[\alpha \].
When \[{K_2}Hg{I_4}\]is dissolved in solvent it gives ions as
\[{K_2}Hg{I_4} \to 2{K^ + }(aq) + Hg{I_4}^{2 - }(aq)\]
The number of ions \[\left( n \right) = 3\]
And the degree of dissociation (\[\alpha \]) is given as\[ = 0.40\]
So, we can write an equation as
\[\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_2}Hg{I_4} \to 2{K^ + }(aq) + Hg{I_4}^{2 - }(aq)\\
initial\,moles\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\
moles\,after\,dissociation\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha
\end{array}\]
So, the total number of moles of ions and the undissociated \[{K_2}Hg{I_4}\]molecules\[ = 1 - \alpha + 2\alpha + \alpha = 2\alpha + 1\]
The its van’t Hoff factor \[(i)\]
\[i = \dfrac{{total\,number\,of\,moles\,}}{{initial\,moles}}\]
By Putting the values of total number of moles we get,
\[\begin{array}{l}
i = \dfrac{{2\alpha + 1}}{1}\\
i = \dfrac{{0.80 + 1}}{1}\\
i = 1.80\\
i > 1
\end{array}\]
So, by using van’t Hoff factor \[(i)\],we can also calculate the extent of dissociation or association.
Therefore, the correct option is option (A).
Note:
The Van't Hoff factor is also defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property. Further, as the colligative property is inversely proportional to the molecular mass of the solute, we can also write that the van't Hoff factor is the ratio of normal molecular mass to the abnormal molecular mass.
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