Question

If $\alpha$ is the degree of dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ , the van't Hoff's factor $\left( i \right)$ used for calculating the molecular mass is:
A. $1 + \alpha$
B. $1 - \alpha$
C. $1 + 2\alpha$
D. $1 - 2\alpha$

Answer 1

The Van’t Hoff factor gives the ratio of the concentration of particles obtained upon the dissolution of the substance to the value of the concentration of the substance calculated by mass.

Complete answer:
The chemical formula ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ represents the compound sodium sulphate. Write the balanced chemical equation for the ionization of sodium sulphate.
${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 2{\text{ N}}{{\text{a}}^ + }{\text{ + SO}}_4^{2 - }$
One molecule of sodium sulphate dissociates to produce two sodium cations and one sulphate anion.
Suppose n molecules of sodium sulphate are present out of which $n\alpha$ molecules of sodium sulphate ionize. Here, $\alpha$ is the degree of dissociation of sodium sulphate.
The dissociation of $n\alpha$ molecules of sodium sulphate will give $2n\alpha$ sodium cations and $n\alpha$ sulphate anions.
Out of n molecules of sodium sulphate, $n\alpha$ molecules dissociate and $n\left( {1 - \alpha } \right)$ molecules of sodium sulphate remain undissociated.
Add the number of sodium ions and the number of sulphate anions to obtain the total number of ions obtained by the dissociation of n molecules of sodium sulphate.
Total number of ions produced is $2n\alpha + n\alpha = 3n\alpha$
To the above number, add the number of undissociated sodium sulphate molecules to obtain the total number of particles present after dissociation
$n\left( {1 - \alpha } \right) + 3n\alpha = n\left( {1 + 2\alpha } \right)$
Calculate the van't Hoff factor

$${\text{i = }}\dfrac{{{\text{Total number of particles present after dissociation}}}}{{{\text{Total number of sodium sulphate molecules}}}} \\\ {\text{i = }}\dfrac{{n\left( {1 + 2\alpha } \right)}}{n} \\\ {\text{i = }}\left( {1 + 2\alpha } \right) \\\$$

If $\alpha$ is the degree of dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ , then the vant Hoff's factor $\left( i \right)$ used for calculating the molecular mass is $1 + 2\alpha$

**Hence, the correct option is option (C).

Note:**
The abnormal molecular mass of the solute is either due to association or due to dissociation. In association, two or more solute particles join to form one unit. In dissociation, one molecule gives two or more ions.