`1.1` mole of A mixed with `2.2` mole of B and the mixture is kept in a 1 litre at the equilibrium `0.2` mole of C is formed, then the value of `K_(c)` will be:

Options

A. 0.002
B. 0.004
C. 0.001
D. 0.003

Answer

C

Explanation

`{:(,A,+,2B,hArr,2C,+,D),("Initial conc.",1.1,,2.2,,0,,0),("Eq. conc.",(1.1-0.1),,(12.2-0.2),,0.2,,0.1):}`
`K=((0.2)^(2)(0.1))/((1.0)(2.0)^(2))=(0.004)/(4)=0.001`
`=1xx10^(-3)`