A dilute solution having on mole of solute 'S' in 1 kg solvent with molal elevation constant `K_(b)` Solute undergoes association as follows. `2ShArrS_(2)` The degree of association of solute `alpha` is given by the following expression . `alpha=(1-i)/(1-1//n)` Where, in the number of molecular of solute undergoing association . The degree of association can be given as :
A
`a = ((K_(b)x - /_\T_(b)))/(/_\T_(b)2)`
B
`a = (2(K_(b)x - /_\T_(b)))/(K_(b)x)`
C
`a =2 + (2/_\T_(b))/(K_(b)x)`
D
`a = (/_\T_(b))/(2K_(b)x)`
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The correct Answer is:
b
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