How many milliliters of an aqueous solution of
0.192 M iron(II) sulfate is
needed to obtain 7.68 grams of the salt?
Molar mass of FeSO4 (Fe(II)SO4] = 151.9 g/mol
That means, 151.1 g FeSO4 salt in 1000 mL = 1 M FeSO4 solution .
i.e. 151.9 g in 1000 mL = 1 M FeSO4
So, 7.68 g in 1000 mL = say 'Z' M
so, Z = 7.68 / 151.9
Z = 0.0506 M
That means, 0.0506 M, 1000 mL FeSO4 solution will give 7.68 g of FeSO4 salt
But given solution has 0.192 M of FeSO4
Let us calculate volume for that concentration,
WWe have 0.0506 M = 1000 ml for 7.68 g of FeSO4
So, 0.192 M = say V mL for 7.68 g of FeSO4
So, V = (1000 x 0.192 ) / (0.0506)
V = 3794.5 mL.
Hence 3794.0 mL of 0.192 M aq.FeSO4 solution will give 7.68 g of the salt.
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