Question

In 249.54kg CuSO4.H2O the number of molecules of water of crystallisation is ___ XN (where N=6*10^23)

Answer 1

mole of CuSO4. 5H2O (249.5grams) contains Avogadro number of molecules. here the answer is 4.158 grams.

249540 grams X 4.158 grams =1037587.32

Answer 2

Answer:

In 249.54kg of CuSO₄.H₂O, the number of molecules of water of crystallization is 1405×N$_A$, where N$_A$ = 6 × 10²³.

Explanation:

Given, the mass of the CuSO₄.H₂O = 249.54Kg = 249.54 × 10³ g

The molecular mass of hydrated copper sulphate (CuSO₄.H₂O) :-

CuSO₄.H₂O $= 63.55+32+(4\times 16)+ 18 =177.55\;g$

One molecule of CuSO₄.H₂O  contains one water molecule.

It means 177.55g of CuSO₄.H₂O has Avogadro number of water molecules.

177.55g of CuSO₄.H₂O has water molecules = 6.023 × 10²³ = $N_A$

249.54 × 10³ g of CuSO₄.H₂O has water molecules:-

$=\frac{249.54\times 10^{3}}{177.55}\times N_A$

$=1.405 \times 10^3\times N_A$

$=1405 \times N_A$

Therefore, the number of water molecules of crystallization of 249.54 kg  CuSO₄.H₂O is equal to 1405×N$_A$.

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