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Degree of dissociation (alpha) alpha a...

Degree of dissociation `(alpha)`
`alpha` are the number of moles which are dissociating from 1 mole of given reactants and gas density measurements can be used to determine the degree of dissociatin. Let us take a general case where one molecule of a substance A splits up into n molecules of A(g) on heating i.e.,
`A_(n)(g)hArrnA(g)`
`t=0a`
`t =t_(eq)a-x nx " "alpha=x/aimpliesx=a alpha`
`a-a alpha n a alpha`
Total number of Moles `=a-a alpha +n a alpha`
`=[1+(n-1)alpha]a`
Observed molecular weight of molar mass of the mixture
`M_("mixture")=(M_(A_(n)))/([1+(n-1)alpha]),M_(A_(n))=` Molar mass of `A_(n)`
A sample of mixture A(g), B(g)and C(g) under equlibrium has a mean molecular weight (observed) of 80.
The equlibrium is
`A(g)hArrB(g)+C(g)`
(Mol wt =100)`" "` (Mol. wt=60)`" "` (Mol. wt=40) Calculate the Degree of dissociation for given reaction.

A

`0.25`

B

`0.5`

C

`0.75`

D

`0.8`

Text Solution

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A
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Degree of dissociation (alpha) alpha are the number of moles which are dissociating from 1 mole of given reactants and gas density measurements can be used to determine the degree of dissociatin. Let us take a general case where one molecule of a substance A splits up into n molecules of A(g) on heating i.e., A_(n)(g)hArrnA(g) t=0a t =t_(eq)a-x nx " "alpha=x/aimpliesx=a alpha a-a alpha n a alpha Total number of Moles =a-a alpha +n a alpha =[1+(n-1)alpha]a Observed molecular weight of molar mass of the mixture M_("mixture")=(M_(A_(n)))/([1+(n-1)alpha]),M_(A_(n))= Molar mass of A_(n) If the t otal mass of the mixture in question (1) is 300 gm, then moles of C(g) present are

The degree of dissociation alpha of a week electrolyte is where n is the number of ions given by 1 mol of electrolyte.

Knowledge Check

  • If alpha is the degree of dissociation of Na_(2)SO_(4) the van't Hoff's factor (i) used for calculating the molecular mass is

    A
    `1+alpha`
    B
    `1-alpha`
    C
    `1+2alpha`
    D
    `1-2alpha`

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