
The Van't Hoff factor calculated from association data is always... than calculated from dissociation data
A. Less
B. More
C. Same
D. More or less
Answer
220.2k+ views
Hint: In the event of solute association, where the observed molecular weight of the solute is larger than the normal, the value of the factor \[i\] is less than unity (that is, \[i < 1\] ), whereas in the case of dissociation, where the observed molecular weight is greater than the normal molecular weight.
Complete step-by-step answer: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, \[i\] . The ratio of the observed or practical colligative quality to the theoretical colligative property is known as the Van't Hoff factor. The fraction of the solute that dissociates in the solution is known as the degree of dissociation. By taking into account both the theoretical and actual dissociation of the electrolyte in the solution, we may determine the relationship between the degree of dissociation and Van't Hoff factor, \[i\] .
The type of solute has no bearing on the properties of the solution; only its proportion or concentration does. The number of particles in a solution increases when polar or ionic solutes are dissolved in polar solvents because they become dissociated. As a result, the computed molar mass, turns out to be less than the theoretical molar mass.
In case of association, 2 molecules will combine to become one associated chemical entity which as higher molecular weight than individual molecules.
The Van't Hoff factor for association is more than 1 and for dissociation is less than 1.
Therefore, the Van't Hoff factor calculated from association data is always less than calculated from dissociation data
Option ‘A’ is correct
Note: Finding the true molar mass of a polar solute becomes challenging at large concentrations of the solute in a solution. As a result, the abnormal molar mass is determined, and the Van't Hoff factor \[i\] is used to determine the true molar mass from it. For dissociation, association, and the non-electrolyte solute, the Van't Hoff factor value is more than one, less than one, and equal to one respectively.
Complete step-by-step answer: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, \[i\] . The ratio of the observed or practical colligative quality to the theoretical colligative property is known as the Van't Hoff factor. The fraction of the solute that dissociates in the solution is known as the degree of dissociation. By taking into account both the theoretical and actual dissociation of the electrolyte in the solution, we may determine the relationship between the degree of dissociation and Van't Hoff factor, \[i\] .
The type of solute has no bearing on the properties of the solution; only its proportion or concentration does. The number of particles in a solution increases when polar or ionic solutes are dissolved in polar solvents because they become dissociated. As a result, the computed molar mass, turns out to be less than the theoretical molar mass.
In case of association, 2 molecules will combine to become one associated chemical entity which as higher molecular weight than individual molecules.
The Van't Hoff factor for association is more than 1 and for dissociation is less than 1.
Therefore, the Van't Hoff factor calculated from association data is always less than calculated from dissociation data
Option ‘A’ is correct
Note: Finding the true molar mass of a polar solute becomes challenging at large concentrations of the solute in a solution. As a result, the abnormal molar mass is determined, and the Van't Hoff factor \[i\] is used to determine the true molar mass from it. For dissociation, association, and the non-electrolyte solute, the Van't Hoff factor value is more than one, less than one, and equal to one respectively.
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