One "mole" of `NH_(4)HS(s)` was allowed to decompose in a `1-L` container at `200^(@)C`. It decomposes reversibly to `NH_(3)(g)` and `H_(2)S(g). NH_(3)(g)` further undergoes decomposition to form `N_(2)(g)` and `H_(2)(g)`. Finally, when equilibrium was set up, the ratio between the number of moles of `NH_(3)(g)` and `H_(2)(g)` was found to be `3`.
`NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2)`
`2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2)`
Answer the following:
To attain equilibrium, how much `%` by weight of folid `NH_(4)HS` got dissociated?

Options

A. `19%`
B. `30%`
C. `33%`
D. `15%`

Answer

C

Explanation

Amount of `NH_(4)HS(s)` dissociated amount of `H_(2)S(g)` present at equilibibrium
`:.` Amount of `NH_(4)HS(s)` dissociated `=0.33 "mol" %` (by weight) of `NH_(4)HS(s)` dissociated
`=0.33/1xx100=33%`