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  • A 0.1 M solution of a weak electrolyte,  , has a freezing point depression of <i

A 0.1 M solution of a weak electrolyte, A_{2}B_{2} , has a freezing point depression of 0.18^{0}C . The van't Hoff factor, i, for the electrolyte is 2. Calculate the degree of dissociation, α.

 

Option: 1

0.60


Option: 2

0.50


Option: 3

0.40


Option: 4

0.30


Answers (1)

best_answer

The freezing point depression, \Delta T f, is given by the formula:
\Delta T f=K f \times \text { molality } \times i
Here, i is the van't Hoff factor, which is 2 for the electrolyte A_2 B_2.
b=\text { (moles of solute/mass of solvent in } \mathrm{kg})
For the solvent, K f=1.86^{\circ} \mathrm{C} / \mathrm{m} molality,
We know that the degree of dissociation, \alpha, for a weak electrolyte can be calculated as:
\alpha=(i-1) /(i+1)(\Delta T f / K f)
Substituting the given values, we get:
\alpha=(2-1) /(2+1) \times(0.18 / 1.86)=0.60

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Gautam harsolia

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