Question

How many molecules of water of crystallisation are present in 1.648g of copper sulphate

Answer 1

Hello dear.

Here is the answer---

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Chemical Formula of the Copper Sulphate Crystamls = CuSO₄.5H₂O

Molar Mass of Copper Sulphate Crystals = 250 g/mole.

∵ 250 grams of the Copper Sulphate Contains 5 × 18 grams of H₂O
∴ 1.648-------------------------------90/250 grams of Water.


Using the Formula,

 Number of Molecules = (Mass/Molecular Mass) × Nₐ
 = (9/25)/18 ×  6.022 ×  10²³
 = 0.12044 × 10²³
 = 12.044 × 10²¹ Molecules.


Thus, the Number of Molecules of the Water of Crystallization in 1.648 grams of Copper Sulphate Crystals  is 12.044 × 10²¹ Molecules.


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Hope it helps.

Answer 2

The molar mass of hydrated copper (II) sulphate = 249.685g while that of 
                                        dry copper (II) sulphate = 159.609g.

Since, the weight of Hydrated copper sulphate is provided with in our question, we can compare the known data to find the unknown weight of dry copper sulphate here.

Let the molecular weight of dry copper sulphate be 'X'. 

        (molar mass of dry CuSO4) / (molar mass of hydrated CuSO4)   =
 (unknown weight of dry CuSO4)/ (Given weight of hydrated CuSO4)

    159.6/249.7 = X/1.648

    X= (159.6*1.648)/249.7 = 1.053

Therefore, the mass of water molecules in hydrated CuSO4= Total weight - X
                                                                                           =   1.648 - 1.053
                                                                                           =      0.595g

Now, Molar mass of 1 H2O molecule = 18g
          
Applying Unitary method,
              
              18g H2O = 1molecule
                1g H2O = (1/18 )molecule
         0.595g H2O = (1/18) * 0.595 molecules
                             =  0.033 molecules

Ans:-  0.033 molecules of H2O are present in 1.648g of CuSO4.