Question

The value of vant hoff factor kcl if it is dissociated 92% will be

Answer 1

$\bigstar$ Answer:

$\checkmark$ Given:

For Dissociation KCl

⇒ Degree of Dissociation ( α ) = 92 %

$\checkmark$ To Find:

⇒ Value of Van't Hoff Factor ( $\bold{i}$ )

$\checkmark$ Solution:

We know that,

Van't Hoff Factor ( $\bold{i}$ )

$\implies \sf i=\dfrac{Normal \ molar \ mass}{Observed \ molar \ mass}$

$\implies \sf i=\dfrac{Obeserved \ colligative \ property}{Normal \ colligative \ property}$

Important Point to Note:

For Solutes that show Dissociation

The Van't Hoff Factor  $\sf i > 1$

Degree of Dissociation ( α ) =

$\green{\boxed{\sf \alpha = \dfrac{i-1}{n-1} }}$

For Solutes that show Association

The Van't Hoff Factor $\sf i < 1$

Degree of Association ( α ) =

$\green{\boxed{\sf \alpha = \dfrac{i-1}{\dfrac{1}{n} -1} }}$

According to the Question,

We are asked to find,

The Value of Van't Hoff Factor ( $\bold{i}$ ) for Dissociation of KCl

From the above formula for Degree of Dissociation ( α )

$\green{\boxed{\sf \alpha = \dfrac{i-1}{n-1} }}$

We are asked to find Value of Van't Hoff Factor ( $\bold{i}$ )

$\sf \implies \alpha(n-1)=i-1$

$\sf \implies \alpha(n-1)+1=i$

Hence,

$\blue{\boxed{\sf i=1+\alpha(n-1)}}$

Where,

α = Degree of Dissociation = 92 %

α = 0.92

n = Number of Species formed after Dissociation

Here,

KCl on Dissociation

$\sf KCl \longrightarrow K^{+}+Cl^{-}$

The Number of Species formed after Dissociation is 2

Hence,

n = 2

Therefore,

Substituting the above values in the Formula,

We get,

$\implies \sf i=1+0.92(2-1)$

$\implies \sf i=1+0.92(1)$

$\implies \sf i=1+0.92$

$\red{\implies \sf i=1.92}$

Hence,

The Value of Van't Hoff Factor ( i ) for

Dissociation of KCl is 1.92

⇒ i = 1.92