Question

How much amount of c u s o 4.5 h2o required for liberation of 2.5 for gram i do when titrated with k?

Answer 1

Molecular weight of CuSO4.5H2O  

=> 63.5 + 32 + 4X16 + 5X18

=>249.5

To prepare 1N solution

we have to dissolve ,= 249/5/2 gm

                               = 124.75 gm

so to prepare 500 cc ,1N solution, 124.75/2

=>62.375 gm

So to prepare 500 cc 0.5N soluiton

                                                       = 62.375 X 0.5

                                                        =  31.18 gm is required.

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