Question

In the laboratory you dissolve 13.7 g of iron(III) sulfate in a volumetric flask and add...

In the laboratory you dissolve 13.7 g of iron(III) sulfate in a volumetric flask and add water to a total volume of 250 mL.

What is the molarity of the solution?

What is the concentration of the iron(III) cation?

What is the concentration of the sulfate anion?

In the laboratory you dissolve 16.2 g of aluminum acetate in a volumetric flask and add water to a total volume of 125 mL.

What is the molarity of the solution?

What is the concentration of the aluminum cation?

What is the concentration of the acetate anion?

In the laboratory you dissolve 19.7 g of barium nitrate in a volumetric flask and add water to a total volume of 500 mL.

What is the molarity of the solution?

What is the concentration of the barium cation?

What is the concentration of the nitrate anion?

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Answer #1

1)

a)

Molar mass of Fe2(SO4)3,

MM = 2*MM(Fe) + 3*MM(S) + 12*MM(O)

= 2*55.85 + 3*32.07 + 12*16.0

= 399.91 g/mol

mass(Fe2(SO4)3)= 13.7 g

use:

number of mol of Fe2(SO4)3,

n = mass of Fe2(SO4)3/molar mass of Fe2(SO4)3

=(13.7 g)/(3.999*10^2 g/mol)

= 3.426*10^-2 mol

volume , V = 2.5*10^2 mL

= 0.25 L

use:

Molarity,

M = number of mol / volume in L

= 3.426*10^-2/0.25

= 0.137 M

Answer: 0.137 M

b)

[Fe3+] = 2*[Fe2(SO4)3]

= 2*0.137 M

= 0.274 M

Answer: 0.274 M

c)

[SO42-] = 3*[Fe2(SO4)3]

= 3*0.137 M

= 0.411 M

Answer: 0.411 M

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