2.14 g of solid ammonium chloride was heated in a one-litre flask to `277^(@)C`. From the measurement pressure, it was found that 90% of ammonium chloride was dissociated. If to this flask 2.04 g of dry ammonia was added, what would be the percentage dissociation ?

Options

A. `75%`
B. `60%`
C. `30%`
D. `25%`

Answer

D

Explanation

`{:(,NH_(4)Cl(s),hArr,NH_(3)(g),+,HCl(g)),("Initial moles",0.04,,0,,0),("At eqbm.",0.004,,0.036,,0.036):}`
`rArr p(NH_(3)) = (0.036 xx 0.082 xx 550)/(V) = 1.6236` atm, `K_(p) = P(NH_(3)) P(HCl) = 2.6236` atm
Partial pressure due to 2.04 g `NH_(3)` is present initially
`(2.04)/(17) xx 0.082 xx 550 = 5.412` atm, If P is the partial pressure of Hcl at equilibrium then
`K_(p) = 2.636 = p(p+ 5.412) = p^(2) + 5.412 p, rArr p^(2) + 5.412p - 2.636 = 0`
`rArr p = (-5.412 + sqrt((5.412)^(2) + 4 xx 2.636))/(2) = 0.45`
`rArr n(HCl) = (0.45 xx 11)/(0.082 xx 550) = 0.01 rArr` % decomposition of `NH_(4)Cl = (0.01)/(0.04) xx 100 = 25%`