Question

If the observed and theoretical molecular mass of NaCl is found to be 31.80 and 58.50. Then the degree of dissociation of NaCl is:
(A) 83.96%
(B) 8.39%
(C) 90%
(D) 100%

Answer 1

When ionic compounds dissolved in water dissociated into cations and anions. This means one mole of a salt dissolved in water, expect to be one mole of cation and one mole of anion to be released in the solution. If this happens, would be two moles of particles in the solution. But the experimentally determined molar mass is always lower than the true value.

Complete step by step solution:
In some cases, a molar mass i.e. either low or higher than the expected or normal value is called an abnormal molar mass.
Van’t Hoff named scientist introduced factor ‘i’, known as the van’t Hoff factor, which is defined as the ratio of normal molar mass to the abnormal molar mass of the solute or the ratio of observed colligative property to calculated colligative property.
Van’t Hoff factor, i = $\dfrac{normal\text{ }molar\text{ }mass}{abnormal\text{ }molar\text{ }mass}$ --- (1)
Van’t Hoff factor can be used to calculate the extent of dissociation or association in terms of the degree of dissociation, which is defined as the fraction of total substance that undergoes dissociation into ions.
Suppose a molecule of an electrolyte after dissociation gives ‘n’ ions and $\alpha$ is the degree of dissociation with 1 mole of solute at equilibrium.
Number of moles of solute left without dissociation = $1-\alpha$
Moles of ions formed = $n\alpha$
Total number of moles at equilibrium = $1-\alpha +n\alpha$
Van’t Hoff factor, $i=\dfrac{1-\alpha +n\alpha }{1}$
$\alpha =\dfrac{i-1}{n-1}$ --- (2)
Given, observed molar mass of NaCl = 31.80 g/mol
Theoretical molar mass of NaCl = 58.50 g/mol
Then, from equation (1), i = $\dfrac{58.5}{31.8}$ = 1.8396
Degree of dissociation of NaCl, substitute the value of ‘i’ in equation (2),
$\alpha =\dfrac{1.8396-1}{2-1}=0.8396$
Hence, the degree of dissociation of NaCl = 89.96%

The correct answer is option A.

Note: In the case of an association, if the Van’t Hoff factor less than 1 means the observed molar mass is more than the normal molar mass. In the case of dissociation, the Van’t Hoff factor is more than 1 because the observed molar mass has a lesser value. If the Van’t Hoff factor is equal to 1, the solute does not undergo dissociation or association in a solvent.