The correct equation for the degree of association 'α' of an associating solute, 'n' molecules of which undergoes association in solution, is

Question

The correct equation for the degree of association 'α' of an associating solute, 'n' molecules of which undergoes association in solution, is (A) α = (n(i -1)/ 1- n) (B) α = (i(n -1)/ 1 + n) (C) α = (i( 1 + n)/ 1- n) (D) α = (i(n + 1)/ 1- n)

Tardigrade Admin · Accepted Answer

n A Leftrightarrow An 1 mole 1-α (α/n), Total =1-α+(α/n) ∴ i=1-α+(α/n) or 1-i=α-(α/n)=α(1-(1/n)) =α((n-1/n)) or α=(n/(n-1))(i-i)=(n(i-1)/(1-n))

Q. The correct equation for the degree of association $^{'} α^{'}$ of an associating solute, $^{'} n^{'}$ molecules of which undergoes association in solution, is

$α = \frac{n ( i - 1 )}{1 - n}$

$α = \frac{i ( n - 1 )}{1 + n}$

$α = \frac{i ( 1 + n )}{1 - n}$

$α = \frac{i ( n + 1 )}{1 - n}$

$α = \frac{n ( 1 - i )}{1 - n}$

$n A \Leftrightarrow A_{n}$

1 mole

$1 - α \frac{α}{n},$ Total $= 1 - α + \frac{α}{n}$

$∴ i = 1 - α + \frac{α}{n}$ or $1 - i = α - \frac{α}{n} = α (1 - \frac{1}{n})$

$= α (\frac{n - 1}{n})$

or $α = \frac{n}{( n - 1 )} (i - i) = \frac{n ( i - 1 )}{( 1 - n )}$

Q. The correct equation for the degree of association ′α′ of an associating solute, ′n′ molecules of which undergoes association in solution, is

α=1−nn(i−1)​

α=1+ni(n−1)​

α=1−ni(1+n)​

α=1−ni(n+1)​

α=1−nn(1−i)​

nA⇔An​ 1 mole 1−αnα​, Total =1−α+nα​ ∴i=1−α+nα​ or 1−i=α−nα​=α(1−n1​) =α(nn−1​) or α=(n−1)n​(i−i)=(1−n)n(i−1)​