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A sampleof 0.15 g of the compound [Pt(NH_(3))_(x)Br_(y)]^(z+) .z Br^(-) , ignited and heated to decomposition produced 0.0502g of Pt. A second0.15 g sample was dissolved in water and titrated rapidly with 0.01 " M " AgNO_(3) solution . 51.50 mL was requiredto precipitate all theionicbromie . A third 0.15 - g sample was heatedfor two hours on a steam both in a solution to which0.2 moleofAgNO_(3) has been added . this precipitated all the bromide ( not just the free ionic Br^(-) ) as AgBr. The weightof the precipitatedthusproduced was 0.20 g . Find x , y and z(Pt = 195 , ag = 108 , Br = 80 , N = 14 and H = 1) |
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Answer» Solution :`[Pt(NH_(3))_(x)Br_(y)]^(2+) .z BR^(-) overset(Delta) to Pt` Applying PoAC for Pt atoms , ` 1 xx ` mole of the COMPOUND = `1XX ` mole of Pt `(0.15)/M = (0.0502)/195""…(1)` where `M = 195 +17x +80 y + 80 z "" ..(2)` For `0.15`G of the secondsample containing z Br atoms molecule( only the ionic bromide ) , one moleculeof the compound shall combine with z molecules of `AgNO_(3)`to give z molecules of AgBr . Applying mole ratio method to reactants `Z xx ` mole of the compound = `1 xx` mole of `AgNO_(3)` `z = (0.15)/M = (0.01 xx 51.5)/1000 ""...(3)` For the 0.15 g of the thirdsample , all bromine atoms `(y+z)`in the compound combine with `AgNO_(3)` to give `(y+z)` molecules of AgBr , Applying mol ratio method , `(y+z) xx` mole of the compound = `1 xx ` mole of AgBr ` (y+z) xx (0.15)/M = 1 xx (0.20)/188 "" ...(4)` Solving equations (1) , (2) ,(3) and (4) we get , x = 4 , y = 2 and z = 2 . |
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