\[{K_2}Hg{I_4}\]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
Verified
473.1k+ views
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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
10 examples of friction in our daily life
The correct order of melting point of 14th group elements class 11 chemistry CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE