Question

1. Calculate the percentage loss of mass of hydrated copper[II] sulphate[CuSO4.5H2O)
when it is completely dehydrated. (CuSO4.5H20 → CuSO4 + 5H20]
[At. wts. are Cu = 64, S = 32, O = 16, H = 1].​

Answer 1

Explanation:

ANSWER

The molecular weight of water H2O=2(1)+16=18.

The molecular weight of water H2O=2(1)+16=18.When 1 mole of CuSO4⋅5H2O is heated, 5 moles of water are lost.

The molecular weight of water H2O=2(1)+16=18.When 1 mole of CuSO4⋅5H2O is heated, 5 moles of water are lost.The molecular weight of CuSO4⋅5H2O=64+32+4(16)+5(18)=250 g/mol

The molecular weight of water H2O=2(1)+16=18.When 1 mole of CuSO4⋅5H2O is heated, 5 moles of water are lost.The molecular weight of CuSO4⋅5H2O=64+32+4(16)+5(18)=250 g/molMass of water lost =5(18)=90 g

The molecular weight of water H2O=2(1)+16=18.When 1 mole of CuSO4⋅5H2O is heated, 5 moles of water are lost.The molecular weight of CuSO4⋅5H2O=64+32+4(16)+5(18)=250 g/molMass of water lost =5(18)=90 gPercentage loss of mass of CuSO4⋅5H2O=25090×100=36 %.

Answer 2

Answer:

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!

Explanation: thankyou