A molecule M associates in a given solvent according to the equation M leftharpoons (M)n. For a certain concentration of M, the van’t Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is :

Question

A molecule M associates in a given solvent according to the equation M leftharpoons (M)n. For a certain concentration of M, the van’t Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is : (A) 3 (B) 5 (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

vanât Hoff factor (i) and the degree of association are related as below: i=-1-α(1-(1/n)) 0.9=1-0.2(1-(1/n)) On solving, (1-(1/n))=(1/2) (1/n)=1-(1/2)=(1/2) ∴ n=2

Q. A molecule $M$ associates in a given solvent according to the equation $M ⇌ (M)_{n}$ . For a certain concentration of $M$ , the van’t Hoff factor was found to be $0.9$ and the fraction of associated molecules was $0.2$ . The value of $n$ is :

$3$

$5$

$2$

$4$

van’t Hoff factor (i) and the degree of association are related as below:
$i = - 1 - α (1 - \frac{1}{n})$
$0.9 = 1 - 0.2 (1 - \frac{1}{n})$
On solving,
$(1 - \frac{1}{n}) = \frac{1}{2}$
$\frac{1}{n} = 1 - \frac{1}{2} = \frac{1}{2}$
$∴ n = 2$

Q. A molecule M associates in a given solvent according to the equation M⇌(M)n​. For a certain concentration of M, the van’t Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is :

3

5

2

4

van’t Hoff factor (i) and the degree of association are related as below: i=−1−α(1−n1​) 0.9=1−0.2(1−n1​) On solving, (1−n1​)=21​ n1​=1−21​=21​ ∴n=2

Q. A molecule $M$ associates in a given solvent according to the equation $M ⇌ (M)_{n}$ . For a certain concentration of $M$ , the van’t Hoff factor was found to be $0.9$ and the fraction of associated molecules was $0.2$ . The value of $n$ is :

$3$

$5$

$2$

$4$

van’t Hoff factor (i) and the degree of association are related as below:
$i = - 1 - α (1 - \frac{1}{n})$
$0.9 = 1 - 0.2 (1 - \frac{1}{n})$
On solving,
$(1 - \frac{1}{n}) = \frac{1}{2}$
$\frac{1}{n} = 1 - \frac{1}{2} = \frac{1}{2}$
$∴ n = 2$