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A solution of NaCl has a Van't Hoff factor of 1.9. If the concentration of NaCl is 0.1 M, calculate

the degree of dissociation \left ( \alpha \right ) of NaCl

Option: 1

0.673


Option: 2

0.790


Option: 3

0.853


Option: 4

0.925


Answers (1)

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Explanation: The Van't Hoff factor (i) for \mathrm{KCl} is 2 since it completely dissociates into \mathrm{K}^{+}and \mathrm{Cl}^{-}ions. The equation to calculate the extent of dissociation is:
i=\frac{1}{1-\alpha+\alpha \times \sqrt{i}}
Substituting the given values, we get:
1.9=\frac{1}{1-\alpha+\alpha \times \sqrt{2}}
Solving for \alpha, we get:
\alpha=0.790

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Pankaj Sanodiya

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