Question

Derive van't hoff solution e​

Answer 1

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Answer 2

Van't hoff factor (i)

It tells us about relationship between normal and colligative properties and abnormal colligative properties.

• If i = 1 ⠀⠀⠀⠀Neither dissociation nor association

• If i > 1 ⠀⠀Dissociation will occur

• If i > 1 ⠀⠀Association will occur

Dissociation of Solute

${\sf Colligative\: Property\:\:\:\:\alpha\:\:\:\:\dfrac{1}{Molecular\: weight\:of\: Solute} }$

Calculation of 'i'

${\sf \:\:\:\:\:\:\:A_xB_y\:\:→\:\:xA^{+y}+yB^{x-}}$

Initially ⠀⠀⠀⠀1 mol ⠀⠀⠀x$\alpha$ ⠀⠀⠀y$\alpha$

After dissociation ⠀⠀(1 - ∝) ⠀⠀x∝ ⠀y∝

${\sf Total\: number\:of\: Solute\: particles=1-\alpha+x\alpha+y\alpha }$

$\implies{\sf 1-\alpha+(x+y)\alpha}$

$\implies{\sf 1-\alpha+n\alpha }$

${\boxed{\sf i=\dfrac{No.\:of\: particles\: after\: dissociation}{No.\:of\: particles\: before\: dissociation} }}$

$\implies{\sf \dfrac{1-\alpha+n\alpha}{1} }$

$\implies{\sf i=1-\alpha+n\alpha}$

Where ∝ is Dissociation

${\boxed{\boxed{\sf \red{\alpha=\dfrac{(i-1)}{(n-1)}}}}}$

• For strong electrolyte

∝ = 1 or 100%

then , n = i

Example :-

NaCl => i = 2

Association of Solute :

Calculation of 'i'

$\large{\sf \:\:\:\:\:nA\:\: →\:\:[A]_n }$

Initially ⠀⠀⠀⠀1 $\huge{→}$⠀⠀0

After association ⠀(1 - ∝) $\huge{→}$ $\sf{\dfrac{\alpha}{n}}$

Total number of Solute particles

$\implies{\sf 1-\alpha+\dfrac{\alpha}{n}}$

${\boxed{\sf i=\dfrac{No.\:of\: particles\:after\: Dissociation}{No.\:of\: particles\:before\: association}}}$

$\implies{\sf i=\dfrac{1-\alpha+\dfrac{\alpha}{n}}{1}}$

$\implies{\sf i=1-\alpha+\dfrac{\alpha}{n}}}$

∝ = degree of association

n = Number of Solute particles which are associated

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

• If ∝ = 100% or 1 or ∝ is not specified ${\boxed{\sf i=\dfrac{1}{n} }}$