Three compounds `A,B` and `C` have empirical formula `CrCI_(3).6H_(2)O` When 1 go of A was kept in a container with dehydrating agent it lost water content and attained constant weight of `0.865g` When 1 g `B` was kept in that vessel it attained a constant weight of `0.932 g` Compund `C` showed no loss in water content .
(a) Find the composition of `A ,B` and `C`
If an excess of aqueous `AgNO_(3)` solution is added to `1g` solution of `A,B` and `C` what amount of `AgCI` will be precipitated in each case .

Explanation

Molecular mass of `CrCI_(3).6H_(2)O`
`=52 + 35.5 xx 3 + 6 xx 18 = 266.5 g`
1 g of A gives `(1-0.865)g`
=0.135 g of `H_(2)O` on dehydration
`:. 266.5 g ` of A gives `=0.135 xx 266.5 g-~36.0 g` of `H_(2)O`
This weight of `36g` shown that on dehydration A gives two moles `H_(2)O ("molecular mass of" H_(2)O) = 1 xx 2+ 16 = 18g)` i.e. two moles of `H_(2)O` are present outside the coordination sphere in the composition of `A` Thus the composition of `A` is given by
`A=[Cr(H_(2)O)_(4)CI_(2)]CI.2H_(2)O`
On similar grounds it can be shown that
`266.5` g of `B` will give `(1-0.932) xx 266.5`
or `18.122 g ` of `H_(2)O`
The weight `(=18.122g)` corresponds to one mole of `H_(2)O` Thus `B` has one `H_(2)O` mole outside the coordination sphere and hence the composition of `B` is give by
`B=[Cr(H_(2)O)_(5)CI]CI_(2).H_(2)O`
Since `C` shows no loss in water content no `H_(2)O` molecule is outside the coordination sphere and hence the comosition of `C` is given by
`C=[Cr(H_(2)O))_(6)]CI_(3)`
(b) Molecular mass of Ag `C1 = 108 + 35.5 = 143.5 g `
Since A i.e, `[Cr(H_(2)O))_(4)CI_(2)]CI.2H_(2)O` on treatment with Ag`NO_(3)` solution givens one one-AgCI mole
`:. 266.5` g of A gives `143.5g` of `AgCI`
:. 1 g of A give `(143.5)/(266.5)g = 0.53` g of AgCI
Since B i e `[Cr(H_(2)O)_(5)CI]CI_(2).H_(2)O `gives two moles of AgCI
`266.5g` of B gives `2 xx 143.5` g of AgCI
:. 1 g of B give `(2 xx 143.5)/(266.5)g = 1.076` g of AgCI
Since C i.e `[Cr(H_(2)O)_(6)]CI_(3)` give three moles of AgCI
266.5 g C give 3 `xx` 1435 g of AgCI
:. 1 g of C give `(3 xx 143.5)/(266.5) g = 1.614` go of `AgCI` .