$PCl X_{5}$ dissociates as $PCl X_{5} PCl X_{3} + Cl X_{2}$ . The equilibrium mixture has a vapour density of $83.4$ (P = 31, Cl = 35.5). The mole fraction of $Cl X_{2}$ in the equilibrium mixture is

A $0.250$
B $0.400$
C $0.200$
D $0.100$

Answer: C

Undissociated VD = 208.5/2 = 104.25. VD(obs) = VD/(1 + alpha), so 1 + alpha = 104.25/83.4 = 1.250 and alpha = 0.250. For 1 mol start: PCl5 = 0.75, PCl3 = Cl2 = 0.25, total 1.25, so x(Cl2) = 0.25/1.25 = 0.200.

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PClX5​ dissociates as PClX5​​PClX3​+ClX2​. The equilibrium mixture has a vapour density of 83.4 (P = 31, Cl = 35.5). The mole fraction of ClX2​ in the equilibrium mixture is

Answer: C

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$PCl X_{5}$ dissociates as $PCl X_{5} PCl X_{3} + Cl X_{2}$ . The equilibrium mixture has a vapour density of $83.4$ (P = 31, Cl = 35.5). The mole fraction of $Cl X_{2}$ in the equilibrium mixture is