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MOLES AND MOLAR CONCENTRATIONS

1. Define the following terms

• Solution

A solution is a homogeneous mixture formed when a solute is dissolved in a solvent. In a solution, the solute particles are dispersed uniformly throughout the solvent. Solutions can be in various states, such as solid solutions (e.g., alloys), liquid solutions (e.g., sugar dissolved in water), or gaseous solutions (e.g., air).

• Solute

A solute refers to a substance that is dissolved in a solvent to form a solution. The solute is typically present in a smaller amount compared to the solvent. It can be a solid, liquid, or gas.

• Solvent

A solvent is a substance that dissolves the solute and forms a solution. It is usually a liquid, but it can also be a gas or a solid in some cases. The solvent is present in a larger quantity than the solute.

• Saturated solution

A saturated solution is a solution in which the solvent has dissolved the maximum amount of solute that it can under given conditions, such as temperature and pressure. At this point, any additional solute added will not dissolve, and instead, it will remain as undissolved particles. The concentration of the solute in a saturated solution is at its solubility limit.

2. Define

(a) Normality: A concentration unit (N); defined as the number of equivalents of solute per liter of solution (e.g., 1 M H2SO4= 2 N H2SO4)

(b) Molarity : The molarity of a solution is defined as the number of moles of solute per one liter of solution. Note that the unit of volume for molarity is liters, not milliliters or some other unit.

Also, note that one liter of solution contains both the solute and the solvent. Molarity, therefore, is a ratio between moles of solute and liters of solution. To prepare laboratory solutions, usually a given volume and molarity are required. To determine molarity, the formula weight or molar mass of the solute is needed.

(c ) ppm

 Parts per million (PPM) is a unit of measurement used when expressing a very dilute concentration level of pollutants in the air, water and other fluids. For example, 1 PPM of ink in water means that in a million mass units of water there would be one mass unit of ink. Thus, PPM refers to one item in a million of anything of the same size. PPM is commonly used in measuring air, water and body fluids pollution. PPM is the mass ratio between a component and a solution. Mathematically, 1 PPM can be expressed as following:

1 PPM = 1 mg/kg

1 PPM = 1 mg/ltr

1 PPM = 0.0001 %

(d) Mass-volume percent solutions  

Are also very common. These solutions are indicated by w/v% and are defined as the grams of solute per 100 milliliters of solution.

Example: 1 g of phenolphthalein in 100 mL of 95% ethyl alcohol is a 1 w/v% solution

(e) Decinormal Solution

The solution that contains 1/10th gram equivalent of solute in one liter of its solution is called decinormal solution.

2  Explain the relationship between Normality and molarity

The number of gram equivalent of solute present in in 1 litre of the solution is called normality. It is represented by N.

Normality=Number of gram equivalent of solute/volume of solution in litre

Normality (N) = Molarity × Basicity of acid

also, Normality (N) = Molarity × Acidity of base

1. Normality Equation

The equation which is used to determine the strength of unknown solution is called normality equation.

Volume of acid × Normality of acid = Volume of Base × Normality of base

3. If 5.00g of sodium chloride is dissolved in exactly 250 cm³of water in a calibrated volumetric flask,

(a) What is the concentration in g/dm³?

     Volume         =    250/1000 = 0.25 dm³

Concentration  =    mass / volume

                        = 5/0.25  

                        = 20 g/dm³

1. What is the molarity of the solution?

A_r(Na) = 23, A_r(Cl) = 35.5,

so M_r(NaCl) = 23 + 35.5 = 58.5

mole NaCl = 5.0/58.5 = 0.08547

volume = 250/1000 = 0.25 dm³

molarity = mol of solute / volume of solvent

Molarity = 0.08547/0.25

                 = 0.342 mol/dm³

c. Calculate the molar concentration  of sulfuric acid from the following data. S.G=1.46

                            %H2SO4= 78%

                              RMM=98

Mass of solution     = 1,000 mL x 1.46 g/mL = 1,460 g

               Mass % = 78    X   1460

                                100    

                            =1138g

              Molar mass of Sulfuric  acid = 98 g/mol

           =    mass x mass %

                   MMHCl 

=     1138g

            98                   =    11.62M

Determine molarity of 37.2% hydrochloric acid (density 1.19 g/mL)

Mass of solution     = 1,000 mL x 1.19 g/mL = 1,190

               Mass % = 37.2 % = 0.372

              Molar mass of hydrochloric acid = 36.4 g/mol

           =mass x mass %

                   MMHCl

                        1,190 g x 0.372

          =       ————————                            

                         36.4 g/mol         

                 = 12.1 moles

  Molarity    = moles/liters

                       = 12.1          

           moles/1 liter = 12.1 M

Calculate the the pH of 1×10⁻⁴M H₂​SO_4​ solution?

Sulphuric acid is a dibasic acid and hence gives  two H+ atoms in it’s solution. Since, it is a strong acid it dissociates completely as:

H2​SO4​⇌2H⁺+SO4²⁻​

          [H+]        = 2 X 0.0001M

          Ph           = −log(2 X 0.0001 )

                   =3.70

1. 4g of sodium hydroxide pellets were disolved  in 1 litre of distilled water . determine the concentration of the solution in

(i) moles per liter

(ii) Parts per million

• 20cm3 of the solution in (a)  was titrated against 25cm3 of sodium hydroxide . determine the concentration of the hydroxide in moles per litre
• Calculate the amount of NaOH required to prepare 0.1Msolution in a 250ml volumetric flask (NaOH=40)
• Describe how solution in (c ) is prepared in the laboaratory

 

2. Calculate the molarity of a solution  containing 4.9g of sulphuric acid in 500cm3 solution (H=1, S=32,0=16)

 

3. Calculate the mass of potassium carbonate required to prepare 0.1M solution in one litre  volumetric flask(K=39,C=12,O=16)

 

4. (a)Determine the molarity of a solution prepared  by dissolving 3.15g of nitric acid in 250ml of water (H=1,N=14, O=16)
5. 20cm3 of a 0.1M NaOH  was neutralized  by 8cm3 of dilute sulphuric acid . calculate the number of moles of the acid .
6. (a)(i) Calculate the amount of KCl  required to prepare 100ppm chloride ions  solution in a 250ml volumetric flask (K=39, Cl=35.5)

(ii)Outline the procedure for laboratory  preparation of dhloride ions  solution in (a)(i)

• Calculate the amount of solution  prepared in (a)(ii) above  required to make 20ml of a 5ppm chloride ion solution
• Determine the number of moles present in 25ml of 0.05Msodium carbonate  solution

 

7. 95g of potassium bromide was dissolved in 400cm³of water.

 

(a) Calculate its molarity. [A_r‘s: K = 39, Br = 80]

moles = mass / formula mass, (KBr = 39 + 80 = 119)

mol KBr = 5.95/119 = 0.050 mol

400 cm³ = 400/1000 = 0.400 dm³

molarity = moles of solute / volume of solution

molarity of KBr solution = 0.050/0.400 = 0.125 mol/dm³

• (b) What is the concentration in grams per dm³?

Concentration = mass / volume, the volume = 400 / 1000 = 0.4 dm³

Concentration = 5.95 / 0.4 = 14.9 g/dm³

8. What mass of sodium hydroxide (NaOH) is needed to make up 500 cm³ (0.500 dm³) of a 0.500 mol dm^-3 (0.5M) solution? [A_r‘s: Na = 23, O = 16, H = 1]

1 mole of NaOH = 23 + 16 + 1 = 40g

molarity = moles / volume, so mol needed = molarity x volume in dm³

500 cm³ = 500/1000 = 0.50 dm³

mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH

therefore mass = mol x formula mass

= 0.25 x 40 = 10g NaOH required

9. How many moles of H₂SO₄are there in 250 cm³ of a 0.800 mol dm^-3 (0.8M) sulphuric acid solution? What mass of acid is in this solution?

                                [A_r‘s: H = 1, S = 32, O = 16]

(a) molarity = moles / volume in dm³, rearranging equation for the sulfuric acid

mol H₂SO₄ = molarity H₂SO₄ x volume of H₂SO₄ in dm³

mol H₂SO₄ = 0.800 x 250/1000 = 0.200 mol H₂SO₄

(b) mass = moles x formula mass

formula mass of H₂SO₄ = 2 + 32 + (4×16) = 98

0.2 mol H₂SO₄ x 98 = 19.6g of H₂SO₄ 

10. A solution of calcium sulphate (CaSO₄) contained 0.500g dissolved in 2.00 dm³ of water. Calculate the concentration in (a) g/dm³, (b) g/cm³ and (c) mol/dm³.

(a) concentration = 0.500/2.00 = 0.250 g/dm³,

then since 1dm³ = 1000 cm³

(b) concentration = 0.250/1000 = 0.00025 g/cm³  

(or from 0.500/2000)

(c) At. masses: Ca = 40, S = 32, O = 64,

formula mass CaSO₄ = 40 + 32 + (4 x 16) = 136

moles CaSO₄ = 0.5 / 136 = 0.00368 mol in 2.00 dm³ of water

concentration CaSO₄ = 0.00368 / 2

11. Describe how a sample of Zinc (II) carbonate can be prepared with Zinc oxide

 

12. Describe how to prepare  250 mL of a 0.30 M NaCl solution.

 A 0.30 M NaCl solution means that there is 0.30 mol NaCl in every liter of solution.

For example:

To make the solution we need to calculate the amount of NaCl to dissolve to make 250 mL of solution.

Start with 250 mL

13. Distinguish between a working solution and stock solution

A stock solution is a large volume of common reagent, such as hydrochloric acid or sodium hydroxide, at a standardized concentration. This term is commonly used in analytical chemistry for procedures such as titrations, where it is important that exact concentrations of solutions are used.

Working solutions are any dilutions that are made using the stock, generally into aqueous solution.

13. Explain the limitation of using  tap  water in preparing  potassium permanagenage standard solution

Any kind of chemical or biochemical experiment demands use of deionized or  distilled water for preparing  all solutions.  Dionized water is highly prefered becouse it is almost completely free of all disolved ions  unlike distilled water which may still be having some dissolved ions . Tap water is not used to prepare standard solutions for titration because of the impurities it contains which could interfere with the results. If you were titating with an acid, the results would be on the high side and if you were titrating with alkaline they would tend to be on the low side because of the minerals likely to be in the tap water.

14. Outline the procedure for standardizing  potassium thocyanate  solution using 0.1M silver nitrate solution

Potassium thiocyanate solution has to be standardized, as it is not possible to prepare and dry KSCN pure enough so that it can be used as a standard substance for solution preparation. The easiest method if the standardization require standardized solution of silver nitrate. As KSCN solution is used for back titration of the excess of AgNO₃, when we need to standardzie KSCN solution we usually have standardized silver nitrate solution ready.

Reaction taking place is

             AgNO₃ + KSCN → AgSCN↓ + KNO₃

Procedure to follow:

• Pipette 25 mL aliquot of about 0.1M AgNO₃solution into 250mL erlenmayer flask.
• Add 50 mL of distilled water.
• Add 1 mL of 10% FeNH₄(SO₄)₂solution.
• Titrate with potassium thiocyanate till the first visible color change.

15. Explain why  iodine solution is prepared using potassium iodide solution

Molecular iodine is only slightly soluble in water. Adding potassium iodide to iodine in water rapidly converts the molecular iodine (I2) to the much more soluble triiodide ion (I3-).

An equimolar solution of iodine and potassium iodide is called Lugol’s solution. The iodine in Lugol’s solution is mostly in the form of triiodide ion, in equilibrium with molecular iodine. In short, potassium iodide is used to be able to attain much higher concentrations of iodine in aqueous solution without having to use organic solvents