The number of moles of ions produced by 1 mol of ionic compound = i

Given, m = 0.0020 m

$$\Delta $$T_f = 0^oC – 0.00732^oC = – 0.00732^oC

K_f = – 1.86 ^oC

$$\Delta $$T_f = ik_fm

$$ \Rightarrow $$ i = $${{0.00732} \over {1.86 \times 0.0020}}$$ = 2

Normality = $${{98 \times 1.8 \times 10} \over {49}}$$ = 36 N

N₂ = 0.1 $$ \times $$ 2 = 0.2 N

N₂V₂ = N₁V₁

$$ \Rightarrow $$ 36 $$ \times $$ V = 0.2 $$ \times $$ 1000

$$ \Rightarrow $$ V = $${{0.2 \times 1000} \over {36}}$$ = 5.55 mL

	HX	⇌	H+	+	X-
Initially	1		0		0
At equilibrium	1 - $$\alpha $$		$$\alpha $$		$$\alpha $$

Total moles = 1 – $$\alpha $$ + $$\alpha $$ + $$\alpha $$ = 1 + $$\alpha $$

$$ \therefore $$ i = $${{1 + \alpha } \over 1}$$

Given, $$\alpha $$ = 20% = 0.2

$$ \therefore $$ i = 1 + $$\alpha $$ = 1 + 0.2 = 1.2

$$\Delta $$T_f = ik_f m = 1.2 × 1.86 × 0.5 = 1.12 K

For isotonic solution,

osmotic pressure of urea = osmotic pressure of nonvolatile solute

$${{10} \over {60 \times 1000}}$$ = $${5 \over {m \times 100}}$$

$$ \Rightarrow $$ m = 300 g mol^–1