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In an evaculated closed isolated chamber...

In an evaculated closed isolated chamber at `250^@C` , 0.02 mole `PCl_5` and 0.1 mole `Cl_2` are mixed. `(PCl_5 ltimplies PCl_3+Cl_2)`.At equilibrium density of mixture was 2.48 g/L and pressure was 1 atm. The number of total moles of equilibrium will be approximately :

A

0.012

B

0.022

C

0.039

D

0.045

Text Solution

Verified by Experts

The correct Answer is:
C
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PCl_5 decomposes as PCl_5(g) hArr PCl_3(g)+Cl_2(g) .If at equilibrium , total pressure is P and density of gaseous mixture is d at temperature T then degree of dissociation (alpha) is : (Molecular wt. of PCl_5=M )

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