1
Question
(i) If 5.5 g of a mixture of FeSO4.7H2Oand Fe2(SO4)3.9H2O required 5.4 mL of 0.1 N KMnO4 solution for complete oxidation, then calculate the number of gram moles of hydrated ferric sulphate in mixture.
(ii) The vapour density of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7oC. Calculate the number of moles of NO2 in 100 g of the mixture.
(i) If 5.5 g of a mixture of FeSO4.7H2Oand Fe2(SO4)3.9H2O required 5.4 mL of 0.1 N KMnO4 solution for complete oxidation, then calculate the number of gram moles of hydrated ferric sulphate in mixture.
(ii) The vapour density of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7oC. Calculate the number of moles of NO2 in 100 g of the mixture.
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Solution
The correct option is C (i) 9.52 mmol, (ii) 0.43 mol
(i)
Let x be the number of gram moles of hydrated ferric sulphate in mixture.
The molar mass of hydrated ferric sulphate Fe2(SO4)3.9H2O is 562 g/mol. The mass of hydrated ferric sulphate will be 562x grams.
Total mass of mixture is 5.5 g.
The mass of hydrated ferrous sulphate will be 5.5−562x grams. The molar mass of hydrated ferrous sulphate is 278 g/mol. The number of moles of hydrated ferrous sulphate will be
5.5−562x278
This is equal to the number of moles of ferrous ions that can be oxidised by potassium permanganate.
5.4 mL of 0.1 N potassium permanganate are used.
Hence, number of g eq of potassium permanganate
=0.1geq/L×5.4ml1000ml/L=0.00054 geq. They are also equal to the number of g eq of ferrous ions. They are also equal to number of moles of ferrous ions.
5.5−562x278=0.00054
5.5−562x=0.15012
562x=5.349
x=9.52×10−3 moles =9.52 mmol.
Because 1 mole = 1000 mmol.
(ii) Let 100 g of mixture contains x moles of NO2
The molecular weight of NO2 is 46 g/mol. The mass of x moles will be 46 x grams.
The mixture will contains 100−46x grams of N2O4
N2O4 has molecular weigh of 92 g/mol. The number of moles of N2O4=100−46x92g/mol
The average molecular mass of the mixture is
46x+92(100−46x92)x+100−46x92
=100x+100−46x92
=920092x+100−46x
=920046x+100....(1)
The vapour density of the mixture is 38.3. Its molecular weight is twice its vapour density. It is 2×38.3=76.6 g/mol ..... (2)
(1) = (2)
920046x+100=76.6
120.1=46x+100
46x=20.1
x=0.437 mol.
The number of moles of NO2 is 0.437.
(i)
Let x be the number of gram moles of hydrated ferric sulphate in mixture.
The molar mass of hydrated ferric sulphate Fe2(SO4)3.9H2O is 562 g/mol. The mass of hydrated ferric sulphate will be 562x grams.
Total mass of mixture is 5.5 g.
The mass of hydrated ferrous sulphate will be 5.5−562x grams. The molar mass of hydrated ferrous sulphate is 278 g/mol. The number of moles of hydrated ferrous sulphate will be
5.5−562x278
This is equal to the number of moles of ferrous ions that can be oxidised by potassium permanganate.
5.4 mL of 0.1 N potassium permanganate are used.
Hence, number of g eq of potassium permanganate
=0.1geq/L×5.4ml1000ml/L=0.00054 geq. They are also equal to the number of g eq of ferrous ions. They are also equal to number of moles of ferrous ions.
5.5−562x278=0.00054
5.5−562x=0.15012
562x=5.349
x=9.52×10−3 moles =9.52 mmol.
Because 1 mole = 1000 mmol.
(ii) Let 100 g of mixture contains x moles of NO2
The molecular weight of NO2 is 46 g/mol. The mass of x moles will be 46 x grams.
The mixture will contains 100−46x grams of N2O4
N2O4 has molecular weigh of 92 g/mol. The number of moles of N2O4=100−46x92g/mol
The average molecular mass of the mixture is
46x+92(100−46x92)x+100−46x92
=100x+100−46x92
=920092x+100−46x
=920046x+100....(1)
The vapour density of the mixture is 38.3. Its molecular weight is twice its vapour density. It is 2×38.3=76.6 g/mol ..... (2)
(1) = (2)
920046x+100=76.6
120.1=46x+100
46x=20.1
x=0.437 mol.
The number of moles of NO2 is 0.437.
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