#### Mole Concept-II, Chemistry tutorial

Introduction:

We are familiar that, the mole was stated and the mole concept employed in calculations. The mole concept is specifically helpful in predicting yield and computing yield from experiments. This is very significant for industrial or laboratory procedures that comprise reversible reactions. Yield and yield percent computations let chemists to evaluate the efficiency of chemical processes and look for ways of enhancement where possible and applicable.

Mole Concept in Solutions:

The molar and molal solutions:

A solution which includes one mole of solute in 1.0 dm3 of the solution is termed as a molar solution. A molar solution is made by dissolving one mole of the solute in small amount of the solvent and then makes the solution up to 1.0 dm3 in a standard volumetric flask. A molar solution can as well be made from amount less than or equivalent to one mole of the solute.

Illustration:

It is needed to prepare 250.00 cm3 of 1.0 mol dm-3 solution of glucose (C6 H12 O6).

Calculate:

a) Mole of glucose needed.

b) Mass of glucose needed.

Solution:

Molar mass of glucose = 6(12) + 12(1) + 6(16) g

= 180 g

For a molar solution of glucose 1dm3 includes 1.0 mol

250 cm3 will have = (1.0 x 250)/(1000) = 0.25 mol

a) 0.25 mol glucose is needed

b) Mass of glucose = mol x molar mass

= 0.25 x 180

= 45 g

A molal solution includes 1 mole in 1 kg of the solvent

A molal solution is made by dissolving 1 mole of the solute in 1000g of solvent. The solute and solvent are weighed and mixed in the affirmed proportions. The final volume of solution is immaterial.

Note:

From the statements and processes of preparation, both solutions are not similar. The preparation of solutions in volumetric (that is, titrimetric) analysis is done in the standard volumetric flasks and solution concentrations are represented in mol dm-3

Not all the solutions are molar solutions. The concentration of a solution is computed from the amount of solute and the volume of solution.

Computation of mole and mass in solution:

If the concentration of a solution is represented in mol dm-3, it is possible to compute the amount in moles and grams of the solute in a volume of solution. You should keep in mind that however a solution consists of one concentration, the amount of solute will be dissimilar for different volumes of the solution. A volume of sea water will taste similar whether you test a cup of it or a bucket or a drop. The amount of salt you recover from sea water though based on the volume of sea water evaporated.

Example:

Compute the amount of sodium chloride recoverable from (a) 1.0 dm3 (b) 10 dm3 (c) 3 drums of sea water. (1 drum - 50 dm3). Concentration of sodium chloride in the sea water is 0.43 mol dm-3. Na = 23, Cl = 35.5 Compute the amount in mole and g dm-3.

Solution:

i)

a) 1.0 dm3 consists of 0.43 mol/NaCl.

b) 10.0 dm3 consists of 0.43 x 10 = 4.3 mol NaCl

c) 3 drums = 3 x 50 dm3 = 150 dm3

150.0 dm3 consists of 0.43 x 150 = 64.5 mol NaCl

ii)

a) Amount in g = mole x molar mass

= 0.43 x 58.5

= 25.16g

b) Amount in g = 4.3 x 58.5

= 251.6g

c) Amount = 64.5 x 58.5 g

= 3773.3g

= 3.77 kg

The Mole Concept in Volumetric Analysis:

Determination of concentration of a solution:

The solution of unknown concentration is titrated by a standard solution (a solution of acknowledged concentration).

Example:

25.0 cm3 of 0.10 mol dm-3 sodium hydroxide solution needs 21.5 cm3 ethanoic acid solutions for complete neutralization by employing phenolphthalein indicator. Compute the concentration of the ethanoic acid solution in (a) mol dm-3 and (b) g/dm3. Equation of the reaction is NaOH+ CH3COOH → CH3COONa + H2O

Solution:

Mole ratio NaOH: CH3 COOH = 1:1

Mole NaOH = [Conc (mol dm-3)/1000] x V (cm3)

= [(0.10 x 25)/1000] = 2.5 x 10-3

As mole ratio is 1:1

Mole CH3 COOH = 2.5 x 10-3

2.5 x 10-3 mole CH3COOH is in 21.5 cm3 of solution

Thus, [(2.5 x 10-3)/21.5] x 1000 mol CH3COOH is in 1.0 dm3 of solution.

Concentration (mol dm-3) = 0.116

Concentration (g/dm3) = mol dm-3 x molar mass

= 0.116 x 60

= 6.96 g dm-3

Determination of percentage purity:

The mole concept can be applied in the volumetric analysis for the determination of percentage purity of an acid or an alkali sample.

Illustration:

5.0 g of sodium hydroxide dissolved in the 1.0 dm3 of solution. 25.0cm3 of the solution needs 24.9 cm3 of a 0.10 mol dm-3 trioxonitrate (v) acid for complete neutralization by employing methyl orange indicator. Compute the percentage purity of the sodium hydroxide sample. The equation of reaction is:

NaOH (aq) + HNO3 (aq) → NaNO3 (aq) + H2O (l)

Mole ratio NaOH: HNO3 = 1:1

Mole HNO3/24.9 cm3 = (24.9 x 0.10)/1000

= 2.49 x 10-3

Thus, mole NaOH/25.0 cm3 = 2.49 x 10-3

Mole NaOH/1000 cm3 = (2.49 x 10-3 x 1000)/25

= 0.0996 mol dm-3

g/dm3 (NaOH) = .0996 x molar mass (NaOH)

= .0996 x 40

= 3.984 g

Percentage purity = (Amount computed/Amount dissolved) x 100

= (3.984/50) x 100

= 79.68

Mole concept in solution dilution:

Numerous times it is needed to make a dilute solution from a relatively concentrated solution of the similar substance in the similar solvent. Mole concept allows for computation to know how much solvent should be added to get the required concentration. Dilution becomes the mere way of making dilute solution of common acids that are available commercially as concentrated acids example: H2SO4, HCl and HNO3.

Illustration:

A bottle of commercial tetraoxosulphate (vi) acids reads 98 percent w/w, specific gravity 1.84, molar mass 98 g.

Compute the volume of this commercial acid sample needed to prepare 2.0 dm3 of a 0.05 mol dm-3 solution.

Solution:

98% w/w means 98g H2SO4 in 100g of solution.

98g H2SO4 are in 100 g of commercial acid

98g H2SO4 are in (100/1.84) cm3 of concentrated acid

It is noted that specific gravity = (mass of substance/mass of an equivalent volume of water)

= (mass of substance/volume of substance)

This signifies that specific gravity is numerically similar as density. Remember,

Density = mass/volume

Thus, Volume = mass/density

98g H2SO4 are in (100/1.84) cm3

Thus, [(98 x 1000 x 1.84)/100] H2SO4 are in 1000 cm3

1803.2 g H2SO4 are in 1000cm3

Concentration (mol dm-3) = [(g/dm3)/molar mass]

= 1803.2/98

= 18.4 mol dm-3

Mole of acid needed for the preparation of 2.0 dm3 of 0.05 mol dm-3 solution

= mol dm-3 x volume (dm3)

= 0.05 x 2 = 0.10 mol dm-3

For concentrated acid:

18.4 mol are present in 1000 cm3

0.10 mol will be present = (0.10/18.4) x 1000

= 5.435 cm3

This latter portion of computation can be done by using the formula:

McVc = MdVd

Mc: concentration of commercial acid

Vc: volume of commercial acid

Md: concentration of dilute acid

Vd: volume of dilute acid.

On substituting,

18.4 x Vc= 0.05 x 2000 cm3

Vc = (0.05 x 2000)/18.4 cm3

= 5.435 cm3

It will be noted that McVc = millimoles of concentration acid

MdVd = millimoles of dilute acid

By equating the millimoles, the volume of concentrated acid needed is computed.

The Mole Concept in Electrolysis:

The mole concept and the Faraday:

Throughout electrolysis chemical reactions take place and substances are liberated or discharged at the electrodes. The Faraday is quantity of charge carried via 1 mole (6.02 x 1023) of electrons. A Faraday discharges or liberates 1 mole of a univalent ion like H+

H+ + e → H

A mole H+ will need a mole of electrons to form a mole of H atoms.

Illustration: Compute the charge on an electron given that Faraday's constant is 96500 coulombs.

By definition:

96500 = 6.02 x 1023 x e, here e is the electronic charge

e = 96500/(6.02 x 1023) = 1.603 x 10-19 coulombs

The mole concept and the efficiency of an electrolytic method:

This can be computed by comparing amount of reaction which occurred at the electrode by the quantity of electricity passed. In such computations the Faraday is employed as a base unit of electricity.

Illustration:

4,825 coulombs of electricity passed in the electrolysis procedure. The volume of hydrogen gas discharged was 0.50 dm3 at S.T.P. Compute the efficiency of the electrolysis method. (Take Faraday constant = 96500 coulombs).

(Hint: 1 Faraday of electricity must liberate 1.0g H2 gas). Molar volume of gas = 22.4 dm3 at S.T.P.)

Solution: 1 Faraday = 6.02 x 1023 electrons = 96500 coulombs

For this procedure:

1 Faraday discharges = 1.0g H2 gas

1 Faraday discharges = (1.0/2.0) x 22.4 dm3 H2 at STP

= 11.2 dm3

1 Faraday discharges 11.2 dm3 at STP

.05 Faraday must discharge.05 x 11.2 = 0.56 dm3 at STP

The efficiency of the procedure = Volume H2 liberated/Theoretical volume

= 0.50/0.56

= 0.89

Percentage efficiency = 89 percent

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