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From the experimental facts given below determine the correct structure of solid `CrCI_(3) .6H_(2)O` Solution containing `0.2665g` of `CrCI_(3)6H_(2)` O was passed throuhg cation exchange resin in acid form The acid liberated was found to react completely with `30mL` of `0.10M` Na OH `("Molecular mass of" CrCI_(3).6H_(2)O=266.5)` . |
Answer» Moles of compound = `(0.2665)/(266.5)=10^(-3)"mol es"` Moles of `NaOH = (30 mL xx 0.1 M)/(1000 mL)=3 xx 10^(-3)` Moles of `NaOH` = moles of `H(30 mL xx 0.1 M)/(1000 mL)=3 x 10^(-3)` Let `1` mole of compunds contains n moles of replaceable `CI ^(Θ)` ions `:. 10^(-3)` moles of compund contains `=n xx 10^(-3)` moles of `CI ^(Θ)` ions `:. n xx 10^(-3) = 3 xx 10^(-3)` Therefore there are `3` moles of replaceable `CI^(Θ)` in the compound So `Cr CI_(3)6H_(2)O` is represented as `[Cr (H_(2)O)_(6)]CI_(3)` . |
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