find the percentage of water of crystallization in feso4, 7H2O
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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 %
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 %
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Answer:
above answer is wron correct answer is 45.34%
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