A 1.0 g impure sample containing [Zn(NH3)4CI2] and some inert impurity was treated with 16 ml of 1.0 M NaOH solution, where all the complex was converted into [Na2Zn(OH)4]. Ammonia formed is first boiled off and then the excess of base required 4 ml of 1.0 M HCI solution for complete neutralisation. After the neutralisation the solution is reacted with excess of AgNO3 solution. the mass of AgCI produced (in mg) is x. The value of x10 is.

Give your Answer........

Calculate the millimoles of SeO3(2−​) in solution on the basis of following data: 70 ml of 60/M​ solution of KBrO3​ was added to SeO3(2−​) solution. The bromine evolved was removed by boiling and excess of KBrO3​ was back titrated with 12.5 mL of 25/M​solution of NaAsO2​. The reactions are given below. I. SeO3(2−​) +BrO3(−​) + H+→SeO42−​+Br2​+H2​O II. BrO3(−​) +AsO2(−​) + H2​O→Br(−) +AsO4(3−​)+ H+

SOLUTION:

A six co-ordinate complex of formula CrCl3.6H2O has green colour. One litre of O.1M solution of the complex when treated with excess of AgNO3 gave 28.7 g of white precipitate. The formula of the complex is .... (1) [Cr(H2O)6]CI3 (2) [Cr(H2O)5CI]CI2.H2O (3) [Cr(H2O)4CI2]Cl.2H2O (4) [Cr(H2O)3Cl3].3H2O

SOLUTION: 

Given 28.7 g of white ppt obtained is 28.7/147.5= 0.2 mole Agcl precipitate Since 0.1M complexgives 0.2 mole AgCl means 2Cl− are ionisable or two Cl- ion must be present out of coordination sphere so, complex is [Cr(H2​O)5​Cl]Cl2​.H2​O