1.00 g of `EuCl_(2)`, is treated with excess of aqueous `AgNO_(3)` and all the chlorine is recovered as 1.29 g of AgCl. Calculate the atomic weight of Eu. (Cl = 35.5, Ag = 108)

Explanation

`underset(1g)(Eu)CI_(2)+AgNO_(3) rarr underset(1.29 g)(AgCI)`
Since CI atoms are conserved applying POAC for CI atoms moles of CI in `EuCI_(2)` = moles of CI in AgCI
`2xx ` moles of `EuCI_(2)=1xx` moles of AgCI
`2xx(1)/((x+35.5xx2))=1 xx(1.29)/((108+35.5))`
( where x = atomic weight of Eu )
`:. x= 152.48`