Question

The value of the observed and calculated molecular weight of silver nitrate is $92.64$ and $170,$ respectively. The degree of dissociation of silver nitrate is:
(A) $60$
(B) $83.5$
(C) $46.7$
(D) $60.23$

Answer 1

Hint :As the amount of solute (silver nitrate) present is the same in both the initial and final solutions, we can equate this on both sides. The amount of water added will be attained by equating the strength of the initial solution and its volume with the strength and volume added of the final solution.

Complete Step By Step Answer:
The following chemical reaction is given by; $AgN{O}_{3(aq)}+A{g}_{(aq)}^{+}+N{O}_{3(aq)}.$ Here we can see that there are two by products formed thus $n=2.$
Dissociation in chemistry can be defined as a general process in which molecules or ionic compounds, such as complexes or salts, split or separate into smaller particles such as atoms, ions or radicals, generally in a reversible manner. The extent to which the dissociation occurs is known as the degree of dissociation. The general equation for determining degree of dissociation is given as $\therefore \alpha ={\displaystyle \frac{(i-1)}{(n-1)}}.$
 The degree of dissociation can be defined as the fraction of solute molecules that dissociates. It can also be defined as the generation of current carrying free ions which dissociate from the fraction of the solute at a given temperature. It is usually defined by $\alpha$ and degree of dissociation is given by $i;$
 $VantHoffFactor=i={\displaystyle \frac{normal(mol.wt)}{observed(mol.wt)}}={\displaystyle \frac{170g/mol}{92.64g/mol}}=1.8350$
Now by substituting the values in equation of dissociation we get;
 $\therefore \alpha ={\displaystyle \frac{(i-1)}{(n-1)}}={\displaystyle \frac{(1.8350-1)}{(2-1)}}={\displaystyle \frac{0.8350}{1}}=0.8350$
 $\alpha =0.8350\times 100=83.50$ degree of dissociation of silver nitrate is $83.5$
Therefore, the correct answer is option B, i.e. $83.5$ .

Note :
There is a difference between the degree of dissociation and the dissociation constant. The degree of dissociation is the dissociation of molecules into smaller particles, ions or radicals, the concentration term unit which we used here gives us the weight of the solute (silver nitrate) per unit volume of the solution. Another commonly used concentration unit is molarity, which gives us the number of moles per unit volume.