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Chemistry For Class IX (New Book ) - Chapter No. 6 - Solutions - Multiple Choice Questions And Fill In The Blanks

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Chapter No.6 - Solutions
MCQs And Fill In The Blanks
Text Book Exercise

SECTION- A: MULTIPLE CHOICE QUESTIONS
Tick Mark (✓) the correct answer

1. An alloy is the homogeneous mixture of:
(a) two solid ✓
(b) two liquid
(c) two gases
(d) solid and liquid

2. A saturated solution of KCl on heating becomes:
(a) unsaturated
(b) supersaturated ✓
(c) diluted
(d) all of these

3. If we dissolve sand into the water, then the mixture is said to be:
(a) solution
(b) suspension ✓
(c) colloids
(d) concentrated solution

4. Solubility is usually expressed in grams of the solute dissolved in ________gram of a solvent.
(a) 10 grams
(b) 100 grams ✓
(b) 500 grams
(d) 1000 grams

5. Example of heterogeneous mixture is:
(a) sugar and water
(b) sand and water ✓
(c ) salt and water
(d) ink and water

6. 2 moles of sodium chloride (NaCl) is equal to:
(a) 123 grams
(b) 135 grams
(c) 158 grams
(d) 117 grams ✓

7. Molarity of a solution which is prepared by dissolving 40 g sodium chloride in 500 cm³ of solution is:
(a) 1.4 M
(b) 1.5 M
(c) 1.33 M
(d) 1.37 M ✓

8. 10% (w/w) sugar solution mean that 10 grams of solute dissolved in:
(a) 90g of water ✓
(b) 95g of water
(c) 100g of water
(d) 105g of water

9. An example of true solution is:
(a) solution of starch
(b) solution of soap
(c) ink in water ✓
(d) tooth paste

10. Which solution contain more water:
(a) 1.0 M
(b) 0.75 M
(c) 0.5 M
(d) 0.25 M ✓

11. When a saturated solution is diluted, it change into:
(a) saturated solution
(b) unsaturated solution ✓
(c) concentrated solution
(d) supersaturated solution

12. Butter is example of solution:
(a) gas-liquid
(b) solid-liquid
(c) liquid-solid ✓
(d) gas-solid

13. A solution that contain solid solute into liquid solvent is called:
(a) solids in gas
(b) liquids in solids
(c) solids in solids
(d) solids in liquid ✓

14. What is the particle size in suspension?
(a) 10³ nm
(b) 10² nm
(c) less than 10³nm
(d) greater than 10³ nm ✓

15. Which of the following statements is true?
(a) A solvent is the homogenous mixture of solute and solution.
(b) A solvent is the heterogenous mixture of solute and solution.
(c) A solution is the homogenous mixture of solute and solvent. ✓
(d) A solution is the heterogenous mixture of solute and solvent.

16. Glass of milk, blood and tap water are the examples of:
(a) solution (Colloid) ✓
(b) solute
(c) solvent
(d) None of them

17. The nutrient absorbed by plants from the soil is the example of the:
(a) solvent
(b) solution ✓
(c) Both 'a' & 'b'
(d) None of these

18. Brass is an example of:
(a) solute
(b) solvent
(c) solution (solid) ✓
(d) None of these

19. Brass is:
(a) zinc dissolved in copper ✓
(b) iron dissolved in copper
(c) iron dissolved in zinc
(d) tin dissolved in iron

20. In aqueous solution, water is present in greater amount and termed as:
(a) solution
(b) solvent ✓
(c) solute
(d) None of these

21. The component of a solution which is always present in smaller amount is called the:
(a) solution
(b) solvent
(c) solute ✓
(d) None of these

22. The component of the solution which is present in larger amount is called:
(a) solution
(b) solvent ✓
(c) solute
(d) None of these

23. It is also called as the universal solvent.
(a) Alcohol
(b) Kerosene oil
(c) Mineral oil
(d) Water ✓

24. A solution that cannot dissolve more solute in it at a particular temperature is called a:
(a) saturated solution ✓
(b) unsaturated solution
(c) supersaturated solution
(d) diluted solution

25. A solution that can dissolve more solute than it contained in the saturated solution after heating is called a:
(a) saturated solution
(b) unsaturated solution
(c) supersaturated solution ✓
(d) diluted solution

26. In carbonated drinks, the state of solute is:
(a) gas ✓
(b) liquid
(c) solid
(d) All of these

27. In fog, the state of solvent is:
(a) gas ✓
(b) liquid
(c) solid
(d) All of these

28. Concentration = ____:
(a) mass of solute in gram x volume of solution in dm³
(b) Mass of solute in gram / volume of solution in dm³ ✓
(c) volume of solution in dm³ / Mass of solute in gram
(d) None of these

29. 1 litre = ___:
(a) 1 cm³
(b) 1 ml
(c) 1000 cm³
(d) 1 dm³ ✓

30. Molarity is defined as the number of moles of solute dissolved in:
(a) 1 ml
(b) 1 cm³
(c) 1 dm³ ✓
(d) 1000 cm³

31. Solubility is increased by:
(a) increasing the temperature ✓
(b) decreasing the temperature
(c) leaving the solution for a period of time
(d) All of these

32. 2 moles of water is equal to:
(a) 18 g
(b) 36 g ✓
(c) 56 g
(d) 46 g

33. A solution that contains large amount of dissolved solute is called:
(a) dilute
(b) saturated
(c) supersaturated
(d) concentrated ✓

34. Homogenous mixture of solute and solvent is called:
(a) solvent
(b) solute
(c) solution ✓
(d) suspension

35. The number of ways of representing percent concentration are:
(a) one
(b) two
(c) three
(d) four ✓

36. Solubility is usually expressed in grams of solute per:
(a) 10 grams of solvent
(b) 100 grams of solvent ✓
(c) 1000 grams of solvent
(d) 10000 grams of solvent

37. In colloids, particles are too big to:
(a) dissolve ✓
(b) absorb
(c) crystallize
(d) disappear

38. Suspensions are:
(a) homogenous
(b) heterogeneous ✓
(c) uniform
(d) solutions

15. Write the example of each type of solution.

Solute	Solvent	Example
Solid	Liquid	Salt in water, Sugar in water
Gas	Gas	Air, mixture of hydrogen and helium in water balloon, oxygen in air
Solid	Solid	Brass an alloy (Zinc dissolved in copper), Bronze (tin dissolved in copper )
Liquid	Solid	Amalgam, butter, cheese
Liquid	Gas	Fog( water vapour in air,) liquid air pollutants
Liquid	Liquid	Alcohol dissolve sin water, oil in ether.

Chemistry Practicals For HSC-Part 1 (XI -Science Group) - Titration - General Descripttion - DJ Sindh Government College

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TITRATION

EQUATIONS:
NaOH + HCl ⟶ NaCl + H₂O
2KOH + H₂SO₄ ⟶ K₂SO₄ + 2H₂O
KOH + HNO₃ ⟶ KNO₃ + H₂O
2NaOH + H₂C₂0₄.2H₂0 ⟶ Na₂C₂0₄ + 4H₂0
2KOH + H₂C₂0₄.2H₂0 ⟶ ₂C₂0₄ + 4H₂0
2HCI + Na₂CO₃ ⟶ 2NaCI + CO₂ + H₂O
H₂SO₄ + K₂CO₃ ⟶ K₂SO₄ + CO₂ + 2H₂0
2HNO₃ + K₂CO₃ ⟶ 2KNO₃ + CO₂ + H₂O
H₂SO₄ + Na₂CO₃ ⟶ Na₂SO₄ + CO₂ + H₂O

2KMn0₄ + 3H₂SO₄ ⟶ K₂SO₄ + 3H₂0 + 2MnSO₄ + 5[O]
10FeSO₄ + 5H₂SO₄ + 5[0] ⟶ 5Fe₂(SO₄)₃ + 5H₂0

---

2KMn0₄ + 10FeSO₄ + 8H₂SO₄ ⟶ K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂0

---

2KMn0₄ + 3H₂SO₄ ⟶ K₂SO₄ + 3H₂0 + 2MnSO₄ + 5[O]
5H₂C₂O₄ + 5[0] ⟶ 10CO₂ + 5H₂0

---

2KMn0₄ + 5H₂C₂O₄ + 3H₂SO₄ ⟶ K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂0

---

IONIC EQUATION FOR ACID — BASE
Na⁺ + OH^- + H⁺ + Cl^- ⟶ Na⁺ + Cl^- + H₂O

2K⁺ + 20H^- + 2H⁺ + SO₄^2- ⟶ 2K⁺ + SO₄^2- + 2H₂0

K⁺ + OH^- + H⁺ + NO₃^- ⟶ K⁺ + NO₃^- + H₂0

2Na⁺ + 20H^- + 2H⁺ + C₂0₄^2- ⟶ 2Na⁺ + C₂0₄^2- + 2H₂0

2K⁺ + OH^- + 2H⁺ + C₂0₄^2- ⟶ 2K⁺ + C₂0₄^2- + 2H₂0

2H⁺ + 2Cl^- + 2Na⁺ + CO₃^2- ⟶ 2Na⁺ + 2Cl^- + CO₂ + H₂O

2H⁺ + SO₄^2- + 2K⁺ + CO₃^2- ⟶ 2K⁺ + SO₄^2- + CO₂ + H₂O

2H⁺ + 2NO₃^- + 2K⁺ + CO₃^2- ⟶ 2K⁺ + 2NO₃^- + CO₂ + H₂0

2H⁺ + SO₄^2- + 2Na⁺ + CO₃^2- ⟶ 2Na⁺ + SO₄^2- + CO₂ + H₂O

THEORY
FOR ACID-BASE TITRATION:
Titration is a process in which the fixed volume of a solution taken by means of pipette into conical flask is compared with the volume of solution added from burette upto end point which is obtained by colour change. The purpose of titration is to determine the unknown strength of a solution.
Base is a substance which gives OH^- ion in aqueous solution. When an acid solution reacts completely with base solution or vice versa, a salt is formed.
Neutralization: In acid-base reaction H⁺ ion from an acid and OH^- ion from a base form H₂O, salt is also formed process is called neutralization. This is indicated by end point and is observe by suitable indicator those having different colors in different medium.
Strong acid and strong base ionize completely in aqueous solution while weak acid and weak base ionize only partially in aqueous solution, these form acidic and basic salts which hydrolyse to change the pH.

Na₂CO₃ + H₂O ⟶ NaOH + H₂CO₃

(Basic salt) + (water) (Strong base) (Weak acid)

where Basic (pH > 7)

FOR REDOX TITRATION:
Titration is a process in which the fixed volume of a solution taken by means of pipette Into conical flask is compared with the volume of solution added from burette upto end point which is obtained by colors change. The purpose of titration is to determine the unknown strength of a solution using standard solution.
Redox Titration: The type of titration in which one reactant in oxidized and other reduced are know as redox titration.
Oxidation is a process in which reducing Agent loses electron and absorbs oxygen, so the oxidation number increases.

Fe²⁺ ⟶ Fe³⁺ + e^-

2FeSO₄ + H₂SO₄ + [0] ⟶ Fe₂(SO₄)₃ + H₂O

Reduction is a process in which oxidizing agent gain electron and loses oxygen, so oxidation number decreases

Mn⁷⁺ + 5e^- ⟶ Mn²⁺

2KMnO₄ + H₂SO₄ ⟶ K₂SO₄ 2MnSO₄ + 3H₂0 + 5[O]

Oxidizing strength of KMnO₄ increase in the presence of H₂SO₄, so H₂SO₄ is required in redox titration. KMnO₄ is coloured substance so it acts as indicator, therefore no external indicator is required.
Reaction between KMnO₄ and oxalic acid is very slow because oxalic acid is a covalent organic compound. So the Titration flask is heated to about 60 — 70°C before adding KMnO₄ from the burette.

PROCEDURE:
Strong Base Vs Stron Acid OR Strong Base Vs Weak Acid

1. Wash burette, pipette and conical flask with water.
2. Rinse burette with base and pipette with acid.
3. Do not rinse the conical flask.
4. Fill the burette with base upto Zero mark.
5. Pipette out 10 cm³ of acid into the conical flask.
6. Add one or two drops of phenolphthalein as indicator.
7. Titrate the base in the burette with the acid in conical flask. The burette solution is added drop by drop into the conical flask.
8. Shake the flask after each addition, The end point is indicated by colour change from colourless to light pink.
9. Note the end point.
10. wash the conical flask with water and repeats the same process for at least three times to obtain two similar readings.


Strong Acid Vs Weak Base
1. Wash burette, pipette and conical flask with water.
2. Rinse burette with acid and pipette with base solution.
3. Do not rinse the conical flask.
4. Fill the burette with acid upto Zero mark.
5. Pipette out 10 cm³ of base into the conical flask.
6. Add one or two drops of methyl orange as indicator.
7. Titrate the acid in the burette with the base in conical flask. The burette solution is added drop by drop into the conical flask.
8. Shake the flask after each addition, The end point is indicated by colour change from yellow to light pink.
9. Note the end point.
10. Wash the conical flask with water and repeat the same process for at least three times to obtain two similar readings.


Redox Titration
1. Wash all the apparatus with water.
2. Rinse burette with KMnO₄ and pipette with reducing agent.
3. Do not rinse the conical flask.
4. Fill the burette with base upto Zero mark.
5. Pipette out 10 cm³ of given reducing agent into the conical flask.
6. Add one test tube of dilute H₂SO₄ (In case of oxalic acid as a reducing agent titration flask is heated to about 60 - 70°C before titration).
7. Titrate KMnO₄ in the burette with the reducing agent in conical flask. The burette solution is added drop by drop into the conical flask.
8. Shake the flask after each addition, the end point is indicated by colour change from colourless to light pink.
9. Note the end point.
10. Wash the conical flask with water repeat the same process for at least three times to obtain two similar readings.

OBSERVATION:

S.NO.	Initial reading	Final reading	Difference	Concordant reading
1.	cm3	cm3	cm3	cm3
2.	cm3	cm3	cm3
3.	cm3	cm3	cm3

• Solution in burette = ______ (Strong Acid Or Strong Base)
• Solution in conical flask = ________ (Weak Acid Or Weak Base)
• Indicator used =  _______ (e.g Phenophthalein for Strong Base)
• Normality of standard solution = N₁ = ___ N
• Volume of standard solution = V₁ = 10 ml
• Normality of solution of unknown strength = N₂ = ?
• Volume of solution of unknown strength = V₂ = ?

CALCULATION:
Standard solution Vs Solution of unknown strength

N₁V₁ = N₂V₂

Equivalent mass = Molecular mass / Acidity or Basicity or change in oxidation state
Amount = Normality (N₂) x equivalent mass x volume of solution in cm³ / 1000

RESULT:

• Normality of given ________ solution = ______ N
• Amount of ______ in the given solution = gms / ___ dm³ or ____ cm³

Acidity Of Base:

Base	Acidity	Equivalent mass
NaOH	1	40
KOH	1	56
Na2CO3	2	53
K2CO3	2	69

Basicity Of Acid

Acid	Basicity	Equivalent mass
HCl	1	36.5
H2SO4	2	49
HNO3	1	63
C2H204.2H20	2	63

Oxidizing Agent

Oxidizing agent	No. of electron gained	Equivalent mass
KMnO4	5	31.6

Reducing Agent

Reducing Agent	No. of electron lost	Equivalent mass
FeSO4.7H20	1	278
FeSO4.(NH4)2SO4.6H20	1	392
H2C204.2H20	2	63

Biology For Class IX - Chapter No.1 - Introduction To Biology - Notes By FAW Group Of Education - Hope Secondary School

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CHAPTER 1
INTRODUCTION TO BIOLOGY
Notes By FAW - Hope Secondary School

Special Thanks To Sir Waseem Ahmed Qureshi

Biology For Class X - Chapter No. 1 - Gaseous Exchange - Notes By FAW Group Of Education

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CHAPTER 1: GASEOUS EXCHANGE

Notes By FAW Group Of Education
(VIDEO LESTURES ARE ALSO AVAILABLE ON WHATSAPP 0315-0226373)

KARACHI GRAMMAR SCHOOL
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Biology For Class X - Chapter No. 1 - Gaseous Exchange - Text Book Chapter

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CHAPTER 1: GASEOUS EXCHANGE

Text Book Chapter