Question

the van't hoff factor for bacl² at 0.01 m concentration is 1.98 the percentage dissociation of bacl² at this concentration is​

Answer 1

Answer:

Reaction: BaCl₂ → Ba²⁺ + 2 Cl⁻

We know that, Van't Hoff Factor is defined as:

$\implies i = \displaystyle{\dfrac{\text{Observed Colligative Property}}{\text{Calculated Colligative Property}}}$

According to the question,

At T = 0,

[ BaCl₂ ] = 0.01 M

[ Ba²⁺ ] = 0

[ Cl⁻ ] = 0

Total Moles = 0.01 + 0 + 0 = 0.01 moles

At T = t

[ BaCl₂ ] = 0.01 M - 0.01 a

[ Ba²⁺ ] = 0.01 a

[ Cl⁻ ] = 0.02 a  [Since question has 2 Cl⁻ ]

Total Moles = 0.01 - 0.01 a + 0.01 a + 0.02 a = 0.01 + 0.02 a

(Here, 'a' is the extent of disassociation.)

Now substituting in formula we get,

$\implies i = \dfrac{0.01 + 0.02a}{0.01}$

According to the question, i = 1.98. Substituting the value we get,

$\implies 1.98 = \dfrac{ 0.01 + 0.02 a}{0.01}\\\\\implies 1.98 \times 0.01 = 0.01 + 0.02 a\\\\\implies 0.0198 - 0.01 = 0.02 a\\\\\implies 0.0098 = 0.02 a\\\\\implies a = \dfrac{0.0098}{0.02} = \boxed{0.49}$

Therefore Percentage Dissociation = a × 100% = 49%