1.

4.5 g of PCl_(5)on vapourisation occupied a volume of 1700mL at atmosphere pressure and 227°C temperature. Calculate it: degree of dissociation.

Answer»


Solution :Let the degree of dissociation be `alpha`, then
`underset(1-alpha)(PCl_(5)) to underset(alpha)(PCl_(3)) + underset(alpha)(Cl_(2))`
(M) MOLECULAR weight of `PCl_(5) = 31+5 xx 35.5 = 208.5`
Total number of molecules before dissociation = 1 Total number of molecules ALTER dissociation
`=1- alpha + alpha + alpha =1 + alpha`
Thus each gram mole CHANGES to `(1+ alpha)` gram-mole.
Thus, `PV =W/M (1+alpha) RT`
(V = Volume after dissociation)
or `1 xx 1700/1000 = 4.5/(208.5) (1+ alpha) RT [1700 ml = 1700/1000 L]`
or `alpha = 0.921`


Discussion

No Comment Found

Related InterviewSolutions