Answer:
9
Explanation:
The following data were obtained from the question:
Mass of hydrous Fe2(SO4)3.nH2O = 0.7632 g
Mass of anhydrous Fe2(SO4)3 =
0.5431 g
Molecular weight of Fe2(SO4)3 = 399.91 g/mol
Value of n =..?
Next, we shall determine the mass of H2O in Fe2(SO4)3.nH2O. This can be obtained as follow:
Mass of hydrous Fe2(SO4)3.nH2O = 0.7632 g
Mass of anhydrous Fe2(SO4)3 =
0.5431 g
Mass of H2O =?
Mass of H2O = (Mass of hydrous Fe2(SO4)3.nH2O) – (Mass of anhydrous Fe2(SO4)3)
Mass of H2O = 0.7632 – 0.5431
Mass of H2O = 0.2201 g
Finally, we shall determine the value of n as follow:
Molar mass of Fe2(SO4)3.nH2O = 399.91 + n(2×1 + 16)
= 399.91 + 18n
Mass of Mass of hydrous Fe2(SO4)3.nH2O = 0.7632 g
Molar mass of nH2O = 18n
Mass of H2O = 0.2201 g
nH2o / Fe2(SO4)3.nH2O = mass of H2O / mass of Fe2(SO4)3.nH2O
18n/399.91 + 18n = 0.2201/0.7632
Cross multiply
18n × 0.7632 = 0.2201 (399.91 + 18n)
13.7376n = 88.020191 + 3.9618n
Collect like terms
13.7376n – 3.9618n = 88.020191
9.7758n = 88.020191
Divide both side by 9.7758
n = 88.020191 / 9.7758
n = 9
Thus, the value of n is 9
