One mole of `N_(2)O_(4)` is heated in a flask with a volume of `10dm^(3).` At equilibrium `1.708` mole of `NO_(2)` and 0.146 mole of `N_(2)O_(4)` were found at `134^(@)C.` The equilibrium constnt will be

Options

A. `250 "mol dm"^(-3)`
B. `300 "mol dm"^(-3)`
C. `2 "mol dm"^3`
D. `230 "mol dm"^(-3)`

Answer

C

Explanation

`N_2O_4 hArr 2NO_2`
`{:("Moles at eqb"^m,0.146,1.708),(underset("mol/dm"^3)(Conc).,[(0.146)/10], [(1.708)/10]):}`
`:. K_c =([NO_2]^2)/([N_2O_4])=([(1.708)/(10)]^(2))/([(0.146)/(10)])`
`implies K_c=(2.917264)/(100)xx10/(0.146)`
`implies K_c = (2.917264)/(1.46) = 2 "mol dm"^(-3)`