25g of a sample of ferrous sulphate was dissolved in water containing dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. calculate the percentage of fe204 7h20 in the sample.
My calculation 0.02x 0.1 = 0.002
0.002 x 1000ml x 293g/mol/ 25ml = 23.44
23.44/25g x 100% = 93.76%
the right answer 88.96%
To calculate the percentage of Fe2O4·7H20 in the sample, we can use the concept of titration.
First, let's calculate the number of moles of KMnO4 solution used in the titration:
Volume of KMnO4 solution used = 20 ml
Molarity of KMnO4 solution = N/10 (N/10 is the same as 0.1 M)
Number of moles of KMnO4 used = Molarity x Volume (in liters)
= 0.1 mol/L x 0.02 L
= 0.002 mol
Next, let's determine the number of moles of Fe2O4·7H20 in the 25 ml of the solution:
Number of moles of Fe2O4·7H20 = Number of moles of KMnO4 used (assuming 1:1 reaction)
= 0.002 mol
Now, let's calculate the molar mass of Fe2O4·7H20:
Fe2O4·7H20 consists of two iron atoms (Fe) with a molar mass of 55.85 g/mol each, four oxygen atoms (O) with a molar mass of 16.00 g/mol each, and 7 water molecules (H2O) with a molar mass of 18.02 g/mol each.
Molar mass of Fe2O4·7H20 = (2 x 55.85 g/mol) + (4 x 16.00 g/mol) + (7 x 18.02 g/mol)
= 111.70 g/mol + 64.00 g/mol + 126.14 g/mol
= 301.84 g/mol
Now, let's calculate the mass of Fe2O4·7H20 in the 25g sample:
Mass of Fe2O4·7H20 = Number of moles x Molar mass
= 0.002 mol x 301.84 g/mol
= 0.60368 g
Finally, calculate the percentage of Fe2O4·7H20 in the sample:
Percentage of Fe2O4·7H20 = (Mass of Fe2O4·7H20 / Mass of sample) x 100%
= (0.60368 g / 25 g) x 100%
= 2.415%
Therefore, the correct percentage of Fe2O4·7H20 in the sample is 2.415%. Please double-check your calculations or provide additional information if you believe the percentage should be 88.96%.