1. Determine the amount concentration, in mol/L, of 0.574 moles of sulfuric acid dissolved in a 281 mL solution. 
2. What is the molar concentration, in mol/L, if 45.1 g of ammonium chloride is dissolved to make 271 mL of solution?
3. Calculate the mass, in grams, of solid magnesium sulfate that is required to make 112 mL of a 4.48 mol/L solution.
4. 104 mL of 0.023 mol/L sulfuric acid solution is diluted up to 448 mL. What is the new concentration of the acid, in mol/L?
5. In a magnesium carbonate solution where the carbonate ion concentration is 0.024 mol/L, what is the concentration of the solute, in mol/L?
6. In a 0.17 mol/L aluminum sulfate solution, what is the molar concentration of the sulfate ion, in mol/L?
7. In a 0.26 mol/L iron (III) sulfate solution, what is the molar concentration of the sulfate ion, in mol/L?
8. 5.34 g of iron(III) sulfate was dissolved in a solution of 543 mL. What is the concentration, in mol/L, of the iron ion in solution?

 

Solution:

 

1)

Given, moles of H2SO4 (n) = 0.574 moles

 

Volume of Solution (V) = 281 mL or 0.281 L

 

we need to calculate concentration in mol/L or simply called Molarity ( mol/L ).

 

The formula to calculate Molarity (M) is:

 

Molarity (M) = No. of moles of Solute (n) / Volume of Solution in Litres

 

So, Molarity (M) = 0.574 moles / 0.281 L

 

Or, Molarity (M) = 2.042 mol/L

 

 

So, Concentration in mol/L = 2.042 mol/L

 

 

 

2)

Given,

 

Mass of ammonium chloride (m) = 45.1 gm

 

Volume of Solution (V) = 271 mL = 0.271 L

 

We know, Molar mass of ammonium Chloride = 53.5 gm/mol

 

 

we need to calculate molar concentration or Molarity of Solution in mol/L

 

 

The formula to calculate Molarity is:

 

Molarity (M) = Number of moles of Solute/ Volume of Solution in Litres

 

 

Firstly, we are required to calculate number of moles of solute (Ammonium Chloride).

 

 

The formula to calculate number of moles is:

 

Number of moles

= Given mass / Molecular Mass

 

So,

 

Number of moles of Ammonium Chloride (n)

 

= 45.1 gm / ( 53.5 gm/mol)

 

= 0.843 mol

 

 

So, putting this value and given Volume in Molarity formula:

 

So,

 

Molarity (M) = (0.843 mol) / 0.271 L

 

Or,

 

M = 3.11 mol/L

 

 

So, molar Concentration Ammonium Chloride = 3.111 mol/L,

 

 

 

3)

 

Given, Volume of Solution (V) = 122 mL = 0.122 L

 

Molarity (M) = 4.48 mol/L

 

We know, molar mass of sodium magnesium sulfate [Na2Mg(SO4)2] = 334.47 gm/mol

 

we need to calculate mass of sodium magnesium sulfate

 

We know,

Molarity (M) = Number of moles of Solute/ Volume of Solution in Litres

 

So, 4.48 mol/L = no. of moles / 0.122 L

 

So, moles of sodium magnesium sulfate = 36.72 moles

 

 

We know,

 

number of moles = Given mass / Molecular Mass

 

So, 36.72 moles = Given mass / 334.47 gm/mol

 

Or, Given mass = 12,282.17 gm

 

 

4)

Given, Initial Volume (V1) = 104 mL = 0.104 L

Initial Concentration (M1) = 0.023 mol/L

 

Final Volume (V2) = 448 mL = 0.448 L

 

we need to calculate final concentration M2 (mol/L).

 

Using the Molarity equation:

 

M1 × V1 = M2 × V2

 

Or, M2 = M1 × V1 / V2

 

So, M2 = [ (0.023 mol/L)×(0.104 L) ] / ( 0.448 L)

 

Or, M2 = 0.00533 mol/L

 

So, new concentration = 0.00533 mol/L

 

 

 

5)

Given, concentration of Carbonate ion = 0.024 mol/L

 

we need to calculate concentration of Magnesium Carbonate

 

The dissociation of MgCO3 is:

 

 

MgCO3 → Mg²⁺ + CO₃^2-

 

We can see that moles of Carbonate ion and magnesium carbonate are same.

 

So, concentration of Magnesium Carbonate = Concentration of carbonate ions

= 0.024 mol/L

 

So, Concentration of Magnesium Carbonate = 0.024 mol/L

 

 

6)

Given, molar concentration of Aluminium sulphate = 0.17 mol/L

 

we need to calculate molar concentration of sulfate ions.

 

The dissociation of Aluminium sulphate is:

 

Al₂(SO₄)₃ → 2 Al³⁺ + 3 SO₄^-

 

We can see that

 

1 mole of aluminium sulphate = 3 moles of sulfate ions

 

So,

 

0.17 mole of aluminium sulphate = 0.17 × 3 = 0.51 moles of sulphate ions

 

 

So, concentration of Sulfate ions = 0.51 mol/L

 

 

7)

Given, concentration of iron (III) sulfate = 0.26 mol/L

 

we need to calculate molar concentration of sulfate ions

 

The dissociation of iron (III) sulfate is:

 

Fe₂(SO₄)₃ →   2 Fe³⁺ + 3 SO₄^-

 

 

We can see that

 

1 mole of iron (III) sulphate = 3 moles of sulfate ions

 

So,

 

0.17 mole of iron (III) sulphate = 0.26 × 3 = 0.78 moles of sulphate ions

 

 

So, concentration of Sulfate ions = 0.78 mol/L

 

 

8)

 

Given, mass of iron(III) sulfate = 5.34 gm

 

Volume of Solution = 543 mL = 0.543 L

 

We know, molar mass of iron(III) sulfate = 400 g/mol

 

So, Number of moles of iron(III) sulfate = 5.34 gm / ( 400 g/mol )

 

Or, moles of iron(III) sulfate = 0.01335 moles

 

We know,

 

Molarity (M) = moles of solute / Volume of Solution in Litres

 

So, Molarity (M) = 0.01335 moles / 0.543 L

 

Or, Molarity (M) = 0.0245 mol/L

 

 

The dissociation of iron (III) sulfate is:

 

Fe₂(SO₄)₃ →   2 Fe³⁺ + 3 SO₄^-

 

 

We can see that

 

1 mole of iron (III) sulphate = 2 moles of iron ions

 

So,

 

0.0245 mole of iron (III) sulphate = 0.0245 × 2 = 0.049 moles of iron ions

 

 

So, concentration of iron ions = 0.049 mol/L