Question

What would be the van't Hoff factor for Alcl3, in water if it is 60% dissociated?​

Answer 1

Answer:

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State True or False.

For a solution of AlCl

3

in water, the Van't Hoff factor (i) is greater than 1.

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Solution

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Correct option is A)

For a solution of AlCl

3

in water ,the Van't Hoff factor (i) is greater than 1 due to hydrolysis.

AlCl

3

+3H

2

O→Al(OH)

3

+3Cl

−

The vant Hoff factor is the ratio of actual number of particles to number of particles for no ionization.

As the number of particles increases due to hydrolysis, the vant Hof factor is more than 1.

Hence, the given statement is true.

Answer 2

Answer:

The Vant' hoff factor for AlCl₃ in water is equal to 2.8.

Explanation:

The dissociation of Aluminium chloride in water :-

AlCl₃ $\longrightarrow$  Al³⁺   +   3Cl⁻

Therefore number of ions, n = 4

Van't Hoff factor gives information about the degree to which a substance associates or dissociates in a solution. The Van't Hoff factor is given by the ratio of the concentration of particles formed  to the substance's mass concentration.

The van't hoff factor when one mole of electrolyte dissociates into n ions will be :-

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

where $\alpha$ is degree of dissociation.

Here, $\alpha = 60\% = 0.60$

Therefore, $i= 1+(4-1)(0.6)$

$i= 1+(3)(0.6)$

$i =2.8$

Therefore,  the Van't Hoff factor for AlCl₃ in water with 60% degree of dissociation is equal to 2.8.

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