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0.2 mole of NH4Cl are introduced into an...

0.2 mole of NH4Cl are introduced into an empty container of 10 litre and heated to `327^(@)C` to attain equilibrium as :
`NH_(4)Cl_((s)) to NH_(3(g)) +HCl_(3(g)) +HCl_((g)) , (K_(P)=0.36"atm"^(2))`
The quantity of solid `NH_(4)Cl` left is:

A

0.078 mole

B

0.02 mole

C

0.095 mole

D

0.035 mole

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(,NH_(4)Cl, to , NH_(3), +, HCl),("moles at " t =0, 0.20,,0,,0),("moles of eqm.",(0.20-a),,a,,a):}`
Also, `K_(p) = P_(NH_(3)) xx P_(HCl) =P^(2)`
`P= sqrt(K_(p))= sqrt(0.36) =0.6` atm
Now, `NH_(3)` formed `n=(PV)/(RT) = (0.6 xx 10)/( 0.0821 xx 600) = 0.122 ` mole
= moles of `NH_(4)Cl`
decomposed
`NH_(4)Cl" left"=0.2 -0.122 = 0.078` mole
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