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Question

Pure phosphine originally present at 2.5 atm and 300 K decomposes slowly according to the equation:
4P H3(g)P4(g)+6H2(g)
What is the vapour density of phosphine if it dissociates to the extent of 40%?

A
13.07
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B
22.17
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C
18.07
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D
19.2
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Solution

The correct option is A 13.07
For the reaction:
4P H3(g)P4(g)+6H2(g)

α=Ddd(n1)
α= dissociation constant
D = theortical vapour Density =GMW2
d = Experimetal Vapour Density
n = No. of moles of product formed.
now
for PH3GMW=31+3(1)=34
D=GMH2=342=17
n=74, as 6H2+one P4 isformed for four moles of reactants.
Given α= 0.4.

putting all the values,
0.4=17dd(741)
0.4(3)(d)=(17d)4

d=13.07

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