Question

Determine the percentage of water of crystallization in pure sample (feso4. 7h2o

Answer 1

Now, molar mass of FeSO₄.7H₂O = 278

No. of moles of water = 7

Mass of one mole of water molecules = 18 g

Mass of 7 moles = 18 x 7 = 126

percentage of water of crystallisation = $\frac{126}{278}$ x 100% = 45.3 % (approx)

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Answer 2

Answer:

atomic mass of Fe = 56 g/mol

atomic mass of S = 32 g/mol

atomic mass of O = 4 g/mol

atomic mass of H = 1 g/mol

now,

molar mass of FeSO4.7H2O = 56 + 32 + 4 × 16 + 7 × { 2 × 1 + 16 }

= 56 + 32 + 64 + 126

= 278 g/mol

molar mass of 7H2O = 7 × { 2 ×1 + 16 } =126 g/mol

now,

% of water in FeSO.7H2O ={ molar mass of 7H2O /molar mass of FeSO4.7H2O }× 100

= { 126/278} × 100

= 12600/278

= 45.32 %