A volatile liquid A 2 B 4 ( M . wt . = 200 ) has the following equilibrium at an elevated temperature : A 2 B 4 ( g ) ⇌ A 2 ( g ) + 2 B 2 ( g ) 0.01 gm of A 2 B 4 , at an elevated temp

Question

A volatile liquid A 2 ​ B 4 ​ ( M . wt . = 200 ) has the following equilibrium at an elevated temperature : A 2 ​ B 4 ​ ( g ) ⇌ A 2 ​ ( g ) + 2 B 2 ​ ( g ) 0.01 gm of A 2 ​ B 4 ​ , at an elevated temperature displaced 2.27 mL of dry air at S.T.P. The degree of dissociation of A 2 ​ B 4 ​ under experimental conditions, is :

TestHub · Accepted Answer

Concepts: Degree of dissociation, vapor density, ideal gas law.

Vapor density method: $D / d = 1 + (n - 1) α$ . Here $n = 3$ . $D = 200/2 = 100$ . $d = (0.01 \times 22400) / (2.27 \times 2) = 49.34$ .

$100/49.34 = 1 + (3 - 1) α ⟹ 2.027 = 1 + 2 α ⟹ α = 0.513 \approx 1/2$ .

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