CBSE Class IX (9th) Science | Chapter 2. IS Matter Around Us Pure ? | Lesson Exercises
Question 1. Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.
Answer :Separation techniques for separation of given substance mixtures :-
| Mixtures of Substance | Separation techniques |
| (a) Sodium chloride from its solution in water. | By Evaoparation |
| (b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride | By Sublimation |
| (c) Small pieces of metal in the engine oil of a car. | By Filtration |
| (d) Different pigments from an extract of flower petals. | Using Chromatography |
| (e) Butter from curd. | By Centrifugation |
| (f) Oil from water. | Using separating funnel |
| (g) Tea leaves from tea. | By Filtration |
| (h) Iron pins from sand. | By a Magnet |
| (i) Wheat grains from husk. | By Winnowing |
| (j) Fine mud particles suspended in water. | By Centrifugation |
Question 2. Write the steps you would use for making tea.
Use the words : solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Answer : For making tea, first of all we will take desired cups of water as solvent in a tea pan. Then it is allowed to boil on stove. Tea leaves, as salute is added to it and brewed. While brewing , the colour of water as solvent gets changed to tea colour as soluble part of tea as salute gets dissolve in heated water as solvent. There after, milk is added to it as solute and is further allowed to boil for some times. Insoluble tea leaves as residue are removed by passing the mixture through a tea strainer. Sugar as solute according to need is added to filtrate so obtained and stirred with spoon. Sugar gets dissolve in filtrate. Resulting solution is in the form of Tea, ready for use.
Question 3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below
(results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).
| Substance Dissolved | Temperature in K 283 293 313 333 353 | ||||
Solubility
| |||||
Potassium nitrate Sodium chloride Potassium chloride Ammonium chloride | 21 36 35 24 | 32 36 35 37 | 62 36 40 41 | 106 37 46 55 | 167 37 54 66 |
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
Answer : Mass of potassium nitrate at 313 K per 100 grams of water in given saturated solution of potassium nitrate = 62 Grams
Required mass of potassium nitrate at 313 K per 50 grams of water for a saturated solution of potassium nitrate = 62 / 100 ×50 = 31 grams
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
Answer : A saturated solution of potassium chloride in water at 353 K has a specific solubility.If it is allowed to cool down at room temperature around 293 K, with the decrease in temperature, solubility of potassium chloride in the solution will also become low. It will dissolve less in water and undissolved potassium chloride result in formation of a layer of potassium chloride crystals at the bottom.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
Answer : The solubility of a saturated solution for a given temperature is expressed as Mass of solute in gm per 100 gm of solvent
- The solubility of Potassium nitrate in a saturated solution with water at 293 K = 32 gm of Potassium nitrate per 100 gm of water
- The solubility of Sodium chloride in a saturated solution with water at 293 K = 36 gm of Sodium chloride per 100 gm of water
- The solubility of Potassium chloride in a saturated solution with water at 293 K = 37 gm Potassium chloride per 100 gm of water
- The solubility of Ammonium chloride in a saturated solution with water at 293 K = 35 gm Ammonium chloride per 100 gm of water
Ammonium chloride has the highest solubility at the given temperature of 293 K
(d) What is the effect of change of temperature on the solubility of a salt?
Answer : Solubility of salt gets gets affected with change of temperature. It increases with rise in temperature and decreases with fall in temperature.
Question 4. Explain the following giving examples.
(a) saturated solution
(b) pure substance
(c) colloid
(d) suspension
Answer :
(a) Saturated solution : A solution at any particular temperature, having maximum amount of solute which can be dissolve in it, is called a saturated solution. In other words, at a given temperature no more solute can be dissolved in a saturated solution.
(b) Pure substance : A pure substance contains only one kind of pure matter and its composition is the same throughout. Mixtures are constituted by more than one kind of pure substance.A pure substance cannot be separated into other kinds of matter by any physical process.Whatever the source of a substance may be, it will always have the same characteristic properties. Examples Sugar, sodium chloride.
(c) colloid : A colloid or a colloidal solution is a heterogeneous mixture in which the particles of a colloid are uniformly spread throughout the solution. Due to the relatively smaller size of particles, the mixture appears to be almost homogeneous, for example, milk.
(d) Suspension : A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particles of a suspension are visible to the naked eye.
The particles of a suspension scatter a beam of light passing through it and make its path visible.The solute particles settle down when a suspension is left undisturbed, that is, a suspension is unstable. They can
be separated from the mixture by the process of filtration. When the particles settle down, the suspension breaks and it does not scatter light any more.
be separated from the mixture by the process of filtration. When the particles settle down, the suspension breaks and it does not scatter light any more.
Question 5. Classify each of the following as a homogeneous or heterogeneous mixture.
soda water, wood, air, soil, vinegar, filtered tea.
Answer :
| Homogeneous Mixture | Heterogeneous Mixture |
|
|
Question 6. How would you confirm that a colourless liquid given to you is pure water?
Answer : For a given atmospheric pressure, every liquid has a distinct boiling point. Water in its pure form boils at 100° C or 273 K at 1 atmospheric pressure. If given colourless liquid starts boiling below or above 100° C or 273 K, it is either not pure water or some other liquid. So boiling point of a liquid, is an indicator of its purity at a particular atmospheric pressure.
Question 7. Which of the following materials fall in the category of a "pure substance" ?
(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air.
Answer : A pure substance contains only one kind of pure matter and its composition is the same throughout. Mixtures are constituted by more than one kind of pure substance.A pure substance cannot be separated into other kinds of matter by any physical process.Whatever the source of a substance may be, it will always have the same characteristic properties. Therefor as per the description of pure substance above, the following materials fall in the category of a “pure substance”:
(a) Ice
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
Question 8. Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water.
Answer : As we know a solution is a homogeneous mixture of two or more substances. The component of the solution which is present in larger amount and dissolves the other is called solvent and other component with lesser quantity which gets dissolved in solvent is called the solute.
As per the description above, the following mixtures are solutions:
(b) Sea water
(c) Air
(e) Soda water
Question 9. Which of the following will show "Tyndall effect"?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.
Answer : Scattering of a beam of light passing through a suspension or a colloidal solution is called the Tyndall effect. Both a suspension and a colloidal solution is a heterogeneous mixture in which the particles which may or may not be seen with naked eyes, can scatter a beam of light passing through it.
As per the description above, following will show "Tyndall effect"
As per the description above, following will show "Tyndall effect"
(b) Milk
(d) Starch solution.Question 10. Classify the following into elements, compounds and mixtures.
(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood
Answer : Classification of materials as elements, compounds and mixtures.
| Elements | Compounds | Mixture |
| (a) Sodium (d) Silver (f) Tin (g) Silicon | (e) Calcium carbonate (k) Methane (l) Carbon dioxide | (b) Soil (c) Sugar solution (h) Coal (i) Air (j) Soap (m) Blood |
Question 11. Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle
Answer :
(a) Growth of a plant
(b) Rusting of iron
(d) Cooking of food
(e) Digestion of food
(g) Burning of a candle
