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Van't Hoff Factor and Abnormal Molar Mass MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • van't Hoff factor(i) or Abnormal Colligative Property is considered one the most difficult concept.

  • Calculation of Extent of Dissociation in an Electrolytic Solution is considered one of the most asked concept.

  • 62 Questions around this concept.

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Which one of the following aqueous solutions will exhibit highest boiling point?

Of the following 0.10 aqueous solutions, which one will exhibit the largest freezing point depression?

If \alpha is the degree of dissociation of Na_{2}SO_{4} the Van't Hoff’s factor (i) used for calculating the molecular mass is

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Concepts Covered - 3

van't Hoff factor(i) or Abnormal Colligative Property

If a solute gets associated or dissociated in a solution, the actual number of particles are different from expected or theoretical consideration.

We know, that:

\mathrm{Colligative\: property \propto number \: of \: particles}

Thus, we can say that:

\\\mathrm{i\: =\: \frac{Observed\: value\: of collgative\: property}{Theoretical\: value\: of\: colligative\: property}}\\\\\mathrm{i\: =\: \frac{Observed\: number\: of\: solute\: particles}{Number\: \: of\: particles\: initially\: taken}}

Again, we have:

\mathrm{Colligative\: property \propto\: \frac{1}{molecular\: mass\: of\: solute}}

Thus;

\mathrm{i\: =\: \frac{Theoretical\: molecular\: mass\: of \: solute}{Observed\: molecular\: mass\: of \: solute}}

 

Calculation of Extent of Dissociation in an Electrolytic Solution

Suppose we have the solute A and it dissociates into B. Then the dissociation occurs as follows:

                           \mathrm{A\rightarrow nB}
At time t = 0         1          0

At time t = t       1 - \mathrm{\alpha }      n\mathrm{\alpha }

Thus, at time t = 0, initial number of solute particles = 1
And, at time t = t, observed number of solute particles = 1 - \mathrm{\alpha } + n\mathrm{\alpha }
                                                                               = 1 + (n-1)\mathrm{\alpha }

Thus, we know that:

\mathrm{i\: =\: \frac{observed\: number\: of\: solute\: particles}{initial\: number\: of\: solute\: particles}}

\mathrm{i\: =\: \frac{1+(n-1)\alpha }{1}}

where n = number of solute particles
         \mathrm{\alpha } = Degree of dissociation

NOTE: For strong electrolytes, \mathrm{\alpha } = 1 

Calculation of Extent of Association in an Electrolytic Solution

Suppose we have the solute A and it associates into (A)n. Then the association occurs as follows:

                        \mathrm{nA\rightarrow (A)_{n}}
At time t = 0        1            0

At time t = t       1 - \mathrm{\alpha }       \mathrm{\alpha }/n

Now, initial number of solute particles = 1

And, observed number of solute particles = 1 - \mathrm{\alpha } + \mathrm{\alpha }/n
                                                             = 1 + \mathrm{\alpha }[1/n - 1]

Thus, van't Hoff factor is given as:

\mathrm{i\: =\: 1+\alpha \left [ \frac{1}{n} -1\right ]}
where, \mathrm{\alpha } is the degree of association

Study it with Videos

van't Hoff factor(i) or Abnormal Colligative Property
Calculation of Extent of Dissociation in an Electrolytic Solution
Calculation of Extent of Association in an Electrolytic Solution

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