1ST TERM

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TOPIC: AMOUNT OF SUBSTANCES- MOLE RATIOS, SOLUTIONS, CONCENTRATION TERMS, STANDARD SOLUTIONS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Define mole, molar solution, standard solution, molarity (molar concentration), mass concentration, molar mass.
b. Calculate number of moles, molarity and molar mass
c. Calculate mass concentration as well as mole ratios.
REFERENCE: NEW CERTIFICATE CHEMISTRY FOR SSCE by OSEI YAW ABABIO (NEW EDITION)

CONTENT: (MOLE RATIOS AND SOLUTIONS) AMOUNT OF SUBSTANCES(STANDARD SOLUTIONS)

A mole is the amount of elementary entities (atoms, molecules, ions) of a substance as there are in 12g of carbon-12.
Solute + solvent = solution
NaCl(s) + H2O(l) NaCl(aq)

A solute is a substance that is dissolved in a solvent to form a solution.
A solvent is a substance which dissolves a solute to form a solution.
A solution is formed when a solute is dissolved in a solvent.
A standard solution is a solution of known concentration, e.g. 5g of NaOH in 1dm3 of solution is a standard solution (or 3moles of KOH in 1dm3 solution).
Concentration of any solution is the amount of the solute in a given volume of solution.
If concentration is in moldm-3, then it is called molar concentration or molarity.
Molarity is concentration in moldm-3 or amount in moles of a solute in 1dm3 of solution.
However, if concentration is in gdm-3, then it is called mass concentration.
A molar solution is one which contains one mole of substance in one dm3 of solution (1 moldm3 of solution (1moldm-3 or 1molar or 1M).

EVALUATION:
1. Define solute, solvent and solution
2. Define a standard solution and a molar solution
3. Explain the difference between molarity and mass concentration.

ASSIGNMENT:
1. Define and give the value of Avogadro number.
2. Explain the difference between a standard solution and a molar solution.






CONTENT: FORMULAS EMPLOYED IN CALCULATION IN TITRATION
Number of moles = mass in g( or reacting mass)
Molar mass (gmol-1)
Number of moles = molarity (moldm-3) X volume (cm3)
1000
Concentration (gdm-3) = molarity (moldm-3) X molar mass (gmol-1)
MAVA = nA; MA = molarity (moldm-3) of acid
MBVB nB VA= volume of acid (cm3)
MB= molarity of base (moldm-3)
VB= Volume of base (cm3)
nA= number of moles of acid
nB= number of moles of base

CALCULATIONS: (h= 1, s= 32, o=16, Na=23, C=12)
14.5cm3 of 4.9gdm-3 of H2SO4 completely neutralized 25cm3 of Na2CO3 in a titration experiment according to the equation:
H2SO4(aq) + Na2SO3(aq) Na2SO4(aq) + H2O(l) + CO2(g)
1. Calculate the concentration of H2SO4 in moldm-3.
2. Calculate the concentration of Na2CO3 in moldm-3
3. Calculate the concentration of Na2CO3 in gdm-3

SOLUTION
1. H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + H2O(l) + CO2(g)
Concentration (gdm-3) = molarity (moldm-3) X molar mass (gmol-1)
Molar mass of H2SO4 = 2H +S +4(0) = (2 X 1) +32 + (4 X 16) gmol-1
Molar mass of H2SO4 = (2 + 32 + 64)gmol-1 = 98gmol-1
Molarity of H2SO4 (moldm-3) = concentration (gdm-3)
Molar mass (gmol-1)
= 4.9gdm-3 = 0.05moldm-3
98gmol-1
Therefore: Concentration of H2SO4 in moldm-3 = 0.05moldm-3

2. Concentration of Na2CO3 in moldm-3
H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + H2O(l) + CO2(g)
1 : 1
MAVA = nA ; MA = 0.05dm-3; VA = 14.5cm3; nA=1
MBVB nB MB= ?; VB= 25cm3; nB=1

0.05moldm-3 X 14.5cm3 = 1
MB X 25cm3 1
MB X 25cm3 = 0.05moldm-3 X 14.5cm3
MB = 0.05moldm-3 X 14.5cm 3 = 0.725moldm-3
25cm3 25
MB= 0.029moldm-3
Therefore concentration of Na2CO3 in moldm-3 = 0.029moldm-3

3. Concentration of Na2CO3 in gdm-3
Concentration (gdm-3) = molarity (moldm-3) X molar mass (gmol-1)
Molar mass (gmol-1) of Na2CO3 = 2Na + C + 3(0)
= (2 X 23) + 12 + (3 X 16) = (46 + 12 + 48)gmol-1
Molar mass (gmol-1) of Na2CO3 = 106gmol-1
Concentration (gdm-3) = 0.029moldm-3 X 106gmol-1
= 3.07gdm-3
Therefore concentration of Na2CO3 in gdm-3 = 3.07gdm-3

EVALUATION:
1. Give the formula for number of moles in terms of reacting mass, molar mass and molarity volume.
2. Give the molarity volume relationship between acid and alkali in acid-base reaction/titration.

ASSIGNMENT:
1. Give the formula for expressing the relationship between mass concentration (gdm-3) and molar concentration (i.e. molarity) in moldm-3.






CONTENT: CALCULATIONS IN ACID-BASE REACTIONS
1. 2g of a mixture of sodium hydroxide and sodium chloride (as impurity) were dissolved in 500cm3 of water. If 25cm3 of this solution were neutralized by 21cm3 of 0.1moldm-3 hydrochloric acid, calculate the percentage of the sodium chloride impurity (NaOH = 40; HCl =36.5; Nacl=58.5)

SOLUTION:
2g of mixture is contained in 500cm3of water.
i.e, 500cm3 of water contains 2g of mixture
1000cm3 of water will contain 2g X 1000cm3
500cm3
=4g of mixture
:. concentration of mixture of NaOH and NaCl is 4gdm-3
MAVA= nA; 21cm3 X 0.1moldm-3 = 1
MBVB nB MB X 25cm3 1
MB X 25cm3 = 21cm3 X 0.1moldm-3
MB = 21cm3 X 0.1moldm-3 = 0.084moldm-3
25cm3
MB = 0.084moldm-3
Therefore molarity of NaOH = 0.084moldm-3
But, molar mass of NaOH = 40gmol-1
Concentration (gdm-3) = molarity (moldm-3) X molar mass (gmol-1)
= 0.084moldm-3 X 40gmol-1
Concentration (gdm-3) = 3.36gdm-3 NaOH
Concentration (gdm-3) of impure NaCl=
Concentration (gdm-3) of mixture of NaOH/NaCl - conc (gdm-3) NaOH
= 4gdm-3 - 3.36gdm-3 = 0.64gdm-3
Therefore concentration (gdm-3) of impure NaCl = 0.64gdm-3
Percentage impurity = conc. Of impure NaCl X 100
Total conc. Of mixture NaCl/NaOH
= 0.64gdm-3 X 100 = 16%
4gdm-3
Therefore percentage of the sodium chloride impurity = 16%

EVALUATION:
1. Calculate percentage purity
2. Calculate percentage impurity
3. In a titration experiment, 26cm3 of H2SO4 neutralized 25cm3 of 0.25moldm-3 of KOH
a. Calculate the molarity of H2SO4 in moldm-3
b. Calculate the mass concentration of H2SO4 and also that of KOH
(H=1, S=32, O=16, K=39)

ASSIGNMENT:
1. Explain how to determine if a given sample of a substance is pure.
( by determining its boiling point and melting point and comparing these with the known and established values of the pure form).

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REVISION