Question

How many milliliters of an aqueous solution of
**0.192** M **iron(II) sulfate** is
needed to obtain **7.68** grams of the salt?

Answer #1

Molar mass of FeSO4 (Fe^{(II)}SO_{4}] = 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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