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SS2: CHEMISTRY - 2ND TERM

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  1. Water & Solution I | Week 1
    10 Topics
    |
    1 Quiz
  2. Water, Solution and Solubility | Week 2 & 3
    9 Topics
    |
    1 Quiz
  3. Air | Week 4
    4 Topics
    |
    1 Quiz
  4. Pollution | Week 5
    6 Topics
    |
    1 Quiz
  5. Rate of Chemical Reaction | Week 6 & 7
    6 Topics
    |
    1 Quiz
  6. Energy and Energy Effect I | Week 8 & 9
    7 Topics
    |
    1 Quiz
  7. Energy and Energy Effect II | Week 10 & 11
    6 Topics
    |
    1 Quiz
  8. Chemical Equilibrium | Week 12
    8 Topics
    |
    1 Quiz

Quizzes

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Question 1:

(a) Give two differences between true and false solution 
(b) Write short notes on the following; 
(i) sols
(ii) emulsion
(iii) aerosol
(iv) fog

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Question 2:

(a) Define the following (i) saturated solution (ii) unsaturated solution (iii) supersaturated solution
(b) 13.5g of KNO3 was completely dissolved in 50cm3 of water. Calculate the solubility of KNO3 in moles/dm3[N = 14, O = 16, K = 39]

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Question 3:

The table below was recorded by a group of students during an experimental investigation of solubility of KCl in water

Temp(°c)010203040506070
Solubility (in mol/dm3)3.94.45.05.56.26.77.48.0

(a) Plot the solubility curve for KCl from the result calculate 

(b)The solubility of NaCl at 54°C 

(c)The temperature at which 100cm3 of water will dissolve 5.9moles of KCl to form a saturated solution

(d)The least amount of water which will dissolve 0.5mole of KCl at 46°C

(e)The mass of KCl precipitate when a saturated solution in 1000cm3 of water is cooled from 66°C to 25°C

Question 4:

(a) What are solubility curves 
(i) Give three uses of solubility curves 

(b) 10.50g of potassium trioxonitrate (V) salt saturated 5g of solvent at 15°C calculate the solubility of potassium trioxonitrate(V) salt at that same temperature in
(i) grammes per dm3           
(ii) moles per dm3   

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Question 5:

In an experiment to determine the solubility of a given salt Y, the following data were provided 
Mass of dry empty dish = 7.16g
Mass of dish + saturated solution of salt Y = 17.85g
Mass of dish + salt Y = 9.30g
Temperature of solution = 20°C
Molar mass of salt Y = 100
Density of solution Y = 100gcm-3 
Calculate the solubility of salt Y in
(i) g/dm3 of solution
(ii) Mol/dm3 of solution  

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Lily Thalia
Lily Thalia
2 years ago

Thanks for this topic. I understood a lot. But pls what is the answer for question 3

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Question 1

(a) Give two differences between true and false solution

Answer:

True Solution False Solution
The solution can pass through a filter paper The solution cannot pass through filter paper
The solution does not scatter light rays (doesn’t undergo tyndall effect The solution can scatter light rays (undergoes tyndall effect)

 

(b) Write short notes on the following;

(i) Sols

Answer: This is the dispersion of solid particles in a liquid medium e.g. glue, gelatin and clay

(ii) Emulsion:

Answer:  This is the dispersion of a liquid in another liquid e.g. Milk, hair cream, cod liver oil.

(iii) Aerosol
Answer: An aerosol is the dispersion of fine solid particles or liquid droplets in air or another gas.

(iv) Fog
Answer: Colloidal state of liquid in gas

Question 2

(a) Define the following (i) saturated solution (ii) unsaturated solution (iii) supersaturated solution

Answer:

(i) Saturated solution of a solute at a particular temperature is one which contains as much as it can dissolve at that temperature in the presence of undissolved solute particles

(ii) Unsaturated solution of a solute at a particular temperature is one which will dissolve more solute at that temperature

(iii) Supersaturated solution is one that contains more solute than it can normally dissolve at a given temperature.


(b) 13.5g of KNO3 was completely dissolved in 50cm3 of water. Calculate the solubility of KNO3 in moles/dm3[N = 14, O =16,K =39

Solution

50cm3 of water contains 13.5g

1000cm3 of the solution will contain

⇒ \( \frac{1000cm^3}{50cm^3} \: \times \: \frac{13.5g}{1} \\ \normalsize = \scriptsize 270g/dm^3 \)

Solubility in mols/dm3 = \( \frac{Solubility\: in \: g/dm^3}{Molar\:Mass} \)

Molar mass of KNO3 = 39 + 14 + (16 × 3)

Molar mass of KNO3 = 101g

= \( \frac{270}{101}\) = 2.67 mol/dm3

Question 4

(a) What are solubility curves

Answer:

Solubility curves are graphical representations of the solubility of different solutes in a particular solvent at different temperatures.

(i) Give three uses of solubility curves

Uses of Solubility Curve

Answer:

  • To determine the amount of solid drug that must be dissolved in a prescribed drug mixture.
  • To determine the most suitable solvent and various temperatures to be used for the extraction of chemicals from natural sources
  • In the separation and purification of a mixture of solute by fractional crystallization


(b) 10.50g of potassium trioxonitrate (V) salt saturated 5g of solvent at 15°C calculate the solubility of potassium trioxonitrate(V) salt at that same temperature in

(i) grammes per dm3

Solution:

5g of KNO3 was saturated by 10.5g

1g of KNO3 was saturated by \( \frac{10.5}{5} \)

1000 cm3 was saturated by \( \frac{10.5}{5} \: \times \: \frac{1000}{1} \\ \scriptsize = 2100gdm^{-3} \)

or

Solubility in g/dm3 = \( \frac{mass\:of\:solute}{vol/mass\:of\: solvent}  \: \times \: \frac{1000}{1} \\ = \frac{10.5}{5 } \: \times \: \frac{1000}{1} \\ = \scriptsize 2100 g/dm^3 \)

(ii) moles per dm3     

Solution:

Solubility in mol/dm3  = \(\frac{Solubility\: in \:gdm^{-3}}{molar\:mass} \)

Molar mass = KNO3 = 39 + 14 + (16 x 3) = 101

Solubility in mol/dm3  = \(\frac{2100}{101} \\ \scriptsize = 20.79 mol/dm^3\)

Question 5

In an experiment to determine the solubility of a given salt Y, the following data were provided

Mass of dry empty dish = 7.16g
Mass of dish + saturated solution of salt Y = 17.85g
Mass of dish + salt Y = 9.30g
Temperature of solution = 20°C
Molar mass of salt Y = 100
Density of solution Y = 1.00gcm-3

Calculate the solubility of salt Y in

(i) g/dm3 of the solution

Solution:

Mass of dish + saturated solution of salt Y   17.85g
Mass of empty dish – 7.16g
Mass of saturated solution = 10.69g
     
Mass of dish + salt Y   48.39g
Mass of empty dish – 7.16g
Mass of solvent = 2.14g

Volume of solution = 10.69cm3
Since volume of solution = \( \frac{mass}{density}\)
and density = 1.0g/dm3

10.69cm3 of solution Y contains 2.14g of salt Y
1cm3 of solution Y contains \( \frac{2.14}{10.69}\)
1000cm3 = \( \frac{2.14\: \times \: 1000}{10.69} \\ \scriptsize 200.19 g/dm^3\)

(ii) Mol/dm3 of solution

Solution:

Solubility in mol/dm3 = \( \frac{solubility \: in \: g/dm^3}{molar\:mass\:of\:salt\:Y}\\ = \frac{200.19}{100} \\ \scriptsize = 2.00 mol/dm^3\)

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