Chemistry, asked by groverraunak2006, 7 months ago

Calculate the percentage loss of mass hydrated copper sulphate when it is completely dehydrated [Cu=64,S=32,O=16,H=1]

Answers

Answered by preetsharma2014in
4

Hey Friend,

Hydrated copper (II) sulphate - CuSO4.xH2O

Considering that the Salt is completely hydrated... x=5

Hydrated copper (II) sulphate - CuSO4.5H2O 

Dehydrated copper (II) sulphate - CuSO4

Molecular weights of -

1. CuSO4.5H2O 

   64 + 32 + (16x4) + 5 [2 + 16]

   64 + 32 + 64 + 90

   250 grams

2. CuSO4

   64 + 32 + (16x4)

   64 + 32 + 64

   160 grams

Therefore,

Difference in the molecular weights... 250 - 160 = 90 grams

Percentage loss =      Difference in weights

                             --------------------------------------    x 100

                        Molecular weight of intial compound

Percentage loss =  90 x 100 / 250

                          = 36 %

Therefore, the percentage loss in the mass when hydrated copper sulphate is completely dehydrated is 36 %.

Hope it helps!

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