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Abnormal vapour density of the mixture of PCl5,PCl3, and Cl2 in the following rxn is approximately PCl5 (g)--->PCl3(g) + Cl2 (g) is found to be 80 (P=31,Cl=35.5) Extent of dissociation of PCl5(g) is approximately (1) 80% (2)50% (3)40% (4)30%

Abnormal vapour density of the mixture of PCl5,PCl3, and Cl2 in the following rxn is approximately PCl5 (g)--->PCl3(g) + Cl2 (g) is found to be 80 (P=31,Cl=35.5) Extent of dissociation of PCl5(g) is approximately (1) 80% (2)50% (3)40% (4)30%

Grade:12

1 Answers

AskiitianExpert Pramod-IIT-R
47 Points
11 years ago
Ans (4) is correct PCl5 == PCl3 + Cl2 c(1-a) ca ca total mole= c(1-a)+ca+ca == c(1+a) equivalent molecular weight = c(1-a)/c(1+a)* M.W. of PCl5 + ca/c(1+a)*M.W. of PCl3 +ca/c(1+a)*M.W. of Cl2 ==> equivalent molecular weight = (1-a)/(1+a)*208.5 + a/(1+a)*137.5 +a/(1+a)*71 ==> equivalent molecular weight = 208.5/(1+a) given vapour density = 80 ==> molecular weight= 2*vapour density ==>molecular weight= 2*80=160 So equivalent molecular weight = 208.5/(1+a) = 160 1+a= 208.5/160 = 1.30 a=.30 means 30% where a is equivalent to degree of dissociation(alfa also) Pramod Kumar IITR alumni

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