Question

Van’t Hoff factor for ${\text{SrC}}{{\text{l}}_{\text{2}}}$ at ${\text{0}}{\text{.01}}$M is $1.6$. Percent dissociation of ${\text{SrC}}{{\text{l}}_{\text{2}}}$is:
A. $70$
B. $55$
C. $40$
D. $30$

Answer 1

Hint: To answer this question we should know the formulas of van't Hoff factor for dissociation and association. As the given compound is ionic and it will go under dissociation we will use the van't Hoff factor formula for dissociation. First, we will write the equation for dissociation to determine the number of ions. Then by substituting all values we will determine the degree of dissociation.

Complete step-by-step solution:
The formula to calculate the van't Hoff factor of is as follows:
${\text{i}}\, = 1 + \,{\alpha }\left( {{\text{n}} - 1} \right)$
Where,
$\,{\text{i}}$ is the van't Hoff factor.
${\alpha }$ is the degree of dissociation
${\text{n}}$ is the number of ions produced by dissociation
We will rearrange the formula for the degree of dissociation as follows:
${\alpha } = \dfrac{{{\text{i}} - 1}}{{{\text{n}} - 1}}$
First, we will determine the number of ions produced by dissociation as follows:
${\text{SrC}}{{\text{l}}_{\text{2}}}\, \to {\text{S}}{{\text{r}}^{2 + }} + \,{\text{2}}\,{\text{C}}{{\text{l}}^ - }$
${\text{SrC}}{{\text{l}}_{\text{2}}}$ dissociates into one strontium ion and two chloride ions so, ${\text{SrC}}{{\text{l}}_{\text{2}}}$ produced a total of three ions so, the value of n is $3$.
On substituting $1.6$ for van't Hoff factor and $3$for number of ions produced,
${\alpha } = \dfrac{{1.6 - 1}}{{3 - 1}}$
${\alpha } = \dfrac{{0.6}}{2}$
${\alpha } = 0.3$
So, the degree of dissociation of ${\text{SrC}}{{\text{l}}_{\text{2}}}$is $0.3$.
To convert the degree of dissociation into percent dissociation we will multiply the degree of dissociation with $100$. So,
Percent dissociation = $0.3 \times 100$%
Percent dissociation = $30$%
So, the percent dissociation of ${\text{SrC}}{{\text{l}}_{\text{2}}}$is $30$.

Hence, option (D) is the correct answer for the given question.

Note:The van't Hoff factor is different for dissociation and the association. Here, ${\text{SrC}}{{\text{l}}_{\text{2}}}$ is an ionic compound. On dissolving in water it undergoes dissociation. So, we use the formula given in the above solution. The formula to determine the van't Hoff factor for the association is as follows:
${\text{i}}\, = \,1 - \left( {1 - \dfrac{{\text{1}}}{{\text{n}}}} \right){\alpha }$
Where,
${\alpha }$ is the degree of association
${\text{n}}$is the number of atoms undergoing polymerization.
The van't Hoff factor is less than one for the association and greater than one for dissociation.