if any solute A dimerises in water at 1atm pressure and the boiling point of this solution is 100.52 degree C. if 2moles of A is added to 1kg of water and K_b for water is 0.52 degreeC/ molal, calculate the percentage association of A

1) 50%

2) 30%

3) 25%

4) 100%

Formula for calculating i (vant haff factor) using given data is :
ΔT = i K_b m .....(1)
ΔT = elevation in boiling point = observed boiling point-original boiling point
     = 100.52$°$ C - 100$°$ C​
      = 0.52$°$ C​
K_b = 0.52$°$ C​ kg mol^-1
m = moles of solute in per kg of solvent = 2 mol kg^-1
Putting the values in given equation (1)
i = $\frac{∆\text{T}}{{\text{K}}_b\times m}$
 $$\begin{gathered} =\frac{0.52°C}{{\text{0.52°C kg mol}}^{\text{-1}}{\text{× 2 mol kg}}^{\text{-1}}} \\ =0.5 \end{gathered}$$
 
 Let $\alpha$ be the degree of association.
At equilibrium the concentration of A would be 1-$\alpha$
Concentration of dimerised form will be $\frac{\alpha }{2}$ mole
Total number of moles in solution =  1-$\alpha$ + $\frac{\alpha }{2}$
                                                  = 1-$\frac{\alpha }{2}$
i = total moles in solution
0.5 = 1-$\frac{\alpha }{2}$
$\frac{\alpha }{2}$= 1/0.5
$\alpha$ = 1
It indicates that percentage of association of A is 100 %
and correct option is 4)



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