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:
Assuming the volume due to solid `NH_(4)HS` is negligible what will be the density of the gaseous mixture in the above equilibrium system?

Options

A. `16.83 g L^(-1)`
B. `16.83 g mL^(-1)`
C. `18.415 g L^(-1)`
D. `14.83 g L^(-1)`

Answer

A

Explanation

Amount of the gases present at equilibrium = weight of `NH_(4)HS(s)` dissociated
`:.` Amount of the gases present `=0.33xx51 g`
Volume of the gases = `1L`
`:.` Density of the gaseous mixtures
`=(0.33xx51)/(1) gL^(-1)=16.83 g L^(-1)`