QJEEM21Ch3ccd: JEE Main 2021 Chemistry - Class 12 JEEM21Ch3ccd Solution: A solute A dimerizes in water. The boiling point of a 2 molal…

$$\mathrm{\Delta }$$T_b = Boiling point of the solution $$-$$ Boiling point of the pure solvent

= 100.52 $$-$$ 100

= 0.52

$$\therefore$$ $$\mathrm{\Delta }$$T_b = K_b (iM)

$$\Rightarrow$$ 0.52 = i $$\times$$ 0.52 $$\times$$ 2

$$\Rightarrow$$ i = $$\frac{1}{2}$$

We know,

i = 1 + $$\left(\frac{1}{n}-1\right)\beta$$

here, $$\beta$$ = degree of dimerization

n = number of particle associated

n = 2 for dimerization

$$\therefore$$ $$\frac{1}{2}=1+\left(\frac{1}{2}-1\right)\beta$$

$$\Rightarrow \beta =1$$

$$\therefore$$ % association = 100