19as you know , i = 1+ (y-1)α..........α= degree of dissociation
y= no of moles of products from one mole reactant
for PCl5 , y=2
=> i=1+α
also, y= (normal v.d) /(abnormal v.d.) = 104.16 / 62 = 1.68
=>α = 0.68
Vapour density of PCl5 is 104.16 but when heated to 230°C its VD is reduced to 62. The degree of dissociation of PCl5 at this temp =
(a) 6.8%
(b) 68%
(c) 46%
(d) 64%
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7 Answers
1919this is the (i) factor used in colligative properties and this question is from that chapter itself...what is the strange thing that you see here ?
but, is the answer correct?
19are u a bit cracked , soumi,,,,,what do u mean by "answer is not there".....α=0.68 means % diss =68% .....thats silly !!! dont mind my words,,,just friendly chat