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The reaction 2ICl(g)I2(g)+Cl2(g) is initiated with 0.75MICl(g). If KC=16 the percentage of ICl(g) decomposed at equilibrium is 

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a
20%
b
40%
c
60%
d
80%

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detailed solution

Correct option is D

We have

2ICl(g)I2(s)+Cl2(g) 0.75Mx x x

Hence, Kc=I2Cl2ICl2=x2(0.75x)2=16

Hence, x0.75Mx=4 or 5x=(0.75M)4 or x=0.6M

Percent of IC! decomposed  =0.6M0.75M×100=80%

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