
One mole of a solute A is dissolved in a given volume of a solvent. The association of the solute take place as follows:
\[nA \rightleftharpoons {A_n}\]
If \[\alpha \] is the degree of association of A, the Van't Hoff factor \[i\] is expressed as:
A. $i = 1 - \alpha $
B. $i = 1 + \dfrac{\alpha }{n}$
C. $i = 1 - \alpha + \dfrac{\alpha }{n}$
D. $i = 1$
Answer
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Hint: A solute's impact on associated properties, such as osmotic pressure, relative vapour pressure reduction, boiling-point elevation, and freezing-point depression, is measured by the Van 't Hoff factor \[i\] . Van’t Hoff factor is less than one for association and degree of association increases with greater extent of association.
Complete step-by-step answer:The Van 't Hoff factor , \[i\] , indicates the difference between the concentration of a material determined by its mass and the concentration of particles actually formed when the substance is diffused. The Van’t Hoff factor is virtually 1 for the majority of non-electrolytes dissolved in water.
In the given reaction, \[n\] moles of solute is giving \[{A_n}\] as the product after reaction. Initially reactant is, say, 1 mole and product is 0 mole. At equilibrium, \[\alpha \] moles associate to give $\dfrac{\alpha }{n}$ moles of the product \[{A_n}\] . So reactant now has \[1 - \alpha \] moles of solute.
Now, to find the Van't Hoff factor we add the number of moles of the reactant and the product.
We get, $i = 1 - \alpha + \dfrac{\alpha }{n}$
Till now we used the conceptual approach to get an expression for the Van't Hoff factor. Let us now use a formula based approach to find an expression for the Van't Hoff factor.
If C is the concentration then, $i = \dfrac{{C\left( {1 - \alpha } \right) + C\left( {\dfrac{\alpha }{n}} \right)}}{C}$
Thus we get the same expression: $i = 1 - \alpha + \dfrac{\alpha }{n}$
Option ‘C’ is correct
Note: The van't Hoff is proportional to the number of unique ions in the substance's formula unit for the majority of ionic compounds dissolved in water. Some solvent also associate in solvent due to formation of association such as dimers. The van’t Hoff value for these type of association is less than 1.
Complete step-by-step answer:The Van 't Hoff factor , \[i\] , indicates the difference between the concentration of a material determined by its mass and the concentration of particles actually formed when the substance is diffused. The Van’t Hoff factor is virtually 1 for the majority of non-electrolytes dissolved in water.
In the given reaction, \[n\] moles of solute is giving \[{A_n}\] as the product after reaction. Initially reactant is, say, 1 mole and product is 0 mole. At equilibrium, \[\alpha \] moles associate to give $\dfrac{\alpha }{n}$ moles of the product \[{A_n}\] . So reactant now has \[1 - \alpha \] moles of solute.
Now, to find the Van't Hoff factor we add the number of moles of the reactant and the product.
We get, $i = 1 - \alpha + \dfrac{\alpha }{n}$
Till now we used the conceptual approach to get an expression for the Van't Hoff factor. Let us now use a formula based approach to find an expression for the Van't Hoff factor.
If C is the concentration then, $i = \dfrac{{C\left( {1 - \alpha } \right) + C\left( {\dfrac{\alpha }{n}} \right)}}{C}$
Thus we get the same expression: $i = 1 - \alpha + \dfrac{\alpha }{n}$
Option ‘C’ is correct
Note: The van't Hoff is proportional to the number of unique ions in the substance's formula unit for the majority of ionic compounds dissolved in water. Some solvent also associate in solvent due to formation of association such as dimers. The van’t Hoff value for these type of association is less than 1.
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