Question

At `523 K, 1` litre of partially dissociated `PCI_(5)` at `1` atm weighs ` 2.695g`. Calculate the percentage dissociation of `PCI_(5) "at "523 K`.

Answer 1

Given: `W_(PCI_(5))=2.695G, Volume=`litre, Pressure=`atm, Temp.=523K`
Asked: `alpha=?`.
Formulae: `PV=(W)/(M)RT`
`M_(0)=(wRT)/(PV)`
Substitution & Calculation: `(2.695xx0.0821xx523)/(1xx1)=115.7`
For `PCI(5,)M_(t)=31+5xx35.5=208.5`
`therefore alpha=(M_(t)-M_(0))/(M_(0))=(208.5-115.7)/(115.7)=0.80. alpha=80%`.