When `20 g` of `CaCO_(3)` were put into `10 litre` flask and heated to `800^(@)C`, `35%` of `CaCO_(3)` remained unreacted at equilibrium. `K_(p)` for decomposition of `CaCO_(3)` is :

Options

A. `1.145 atm`
B. `0.145 atm`
C. `2.146 atm`
D. `3.145 atm`

Explanation

`{:(,CaCO_(3(s)),hArr,CaO_((s)),+,CO_(2(g))),("Mole before dissociation",(20)/(100),,0,,0),("Mole after dissociation",((20)/(100)xx(35)/(100)),,((20)/(100)xx(65)/(100)),,((20)/(100)xx(65)/(100))):}`
`:. "Mole of" CO_(2) "formed"=(20xx65)/(10^(4))=1.3xx10^(-1)`
`:' PV=nRT`
`P_(CO_(2))=(1.3xx10^(-1))/(10)xx0.0821xx1073=1.145 atm`
Now `K_(p)=P'_(CO_(2))=1.145 atm`