Is matter around us pure
2.1 Mixture
It is constituted by more than one kind of pure substance.
Types of mixtures-There are two types of mixture:
• Homogeneous mixture- A mixture which has a uniform composition throughout.
Example- Salt /sugar dissolved in water.
• Heterogeneous mixture- A Mixture which contains the non-uniform composition.
Example -A mixture of oil and water.
2.2 Solution
•It is the homogeneous mixture of two or more substance.
Example – Soda water lemon Juice.
•Alloys are the mixture of :
-Two or more metals
-Metal and a non-metal
which cannot be separated into their components by physical methods.
Example- Brass is a mixture of 30% zinc and 70% copper.
• A solution has a :
Solvent –The component of the solution that dissolves the other component present in it.
Solute-The component of the solution that dissolved in the solvent.
•Properties of a solution-
– A solution is a homogeneous mixture.
– Particle size= smaller than 1nm so cannot be seen by naked eyes
– Do not scatter a beam of light on passing through it.
– Cannot be separated from the mixture by the filtration process.
•Concentration of a solution –
The amount of solute present in a given amount of solution.
Here are the three methods for expressing the concentration of a solution:
(a)Mass by the mass percentage of a solution =Mass of a solute/ mass of solution x 100
(b) Mass by volume percentage of a solution = Mass of solute/volume of solution x 100
(c) volume by volume percentage of a solution = volume of solute/ volume of solution x 100
2.3 Suspension
It is a heterogeneous mixture in which solute particles do not dissolve and they remain suspended throughout the medium. We can see these particles with naked eyes.
•Properties of a suspension-
– It is a heterogeneous mixture.
– we can see the particles with our naked eyes.
– it scatters a beam of light passing through it.
-the suspension is unstable that means particles settle down when the suspension is left undisturbed and we can be separated by the process of filtration.
2.4 Colloidal solution
The solution in which solute particles have a size smaller than that of the suspension but larger than that of a solution. We cannot see it by naked eyes. But, these particles can easily scatter a beam of visible light.
•Scattering of a beam of light by a colloidal solution is called Tyndall effect.
Properties of a solid-
-It is a home heterogeneous mixture.
– Size of a particle is between that of solution and suspension particles.
– Particles are big enough to scatter a beam of light passing through it.
– Particles do not settle down when left undisturbed.
-Cannot be separated from the mixture by the process of filtration but have a special technique for the separation.
2.5 Separating the components of a mixture
Here are some methods of separation of components of a mixture-
•Evaporation-We can separate the volatile from its non-volatile solute by the method of Evaporation.
Application-Coloured component from blue-black
•Centrifugation- When the solid particles are very small and pass-through filter paper.
Application-Diagnostic Laboratories for blood and urine test.
– used in dairies and to make butter from cream.
-used in washing machine squeeze out water from wet clothes.
– Separation of cream from milk
•Sublimation- A mixture that contains a volatile component and non-sublimable impurity.
Application- A separate mixture of salt and camphor.
•Chromatography- It is used for separation of those solutes that dissolve in the same solvent.
Application- to separate colours in a dye.
– Pigment from natural colours.
-Drugs from the blood.
•Distillation – Method uses separation of a component of the mixture containing two miscible liquids that boy without decomposition hand having sufficient difference in their boiling point.
Application-separation of acetone and water from their mixture.
•Fractional distillation- Separation of a homogeneous mixture like air into its components.
Method-Air on compressing and cool by increasing pressure and decreasing temperature changes to liquid air. Liquid air is allowed to warm up slowly in a fractional distillation column that gets separated at different Heights.
•Crystallisation – That separates a pure solid in the form of its crystals from a solution.
Application- purification of salt that we get from sea water.
– separation of crystals of alum from the impure sample.
2.6 Physical and chemical changes
Physical changes- Changes occur without a change in composition.
No change in the chemical nature of a substance.
Example- conversion of ice into water
Chemical changes- One substance react with another to undergo a change in chemical
composition.
It brings a change in the chemical properties of matter and we get an entirely new substance.
2.7 Elements and compounds-
ELEMENTS- The basic form of a method that cannot be divided into a simpler substance by a
chemical reaction.
• Robert Boyle was first to use the term element.
• Elements are divided into three categories :
-Metal – Elements that conduct electricity and are sonorous, malleable and ductile.
Example- Gold , Silver
-Nonmetal – Do not conduct heat and electricity and are not lustrous, sonorous and ductile.
Example- oxygen, hydrogen.
-Metalloid – Those elements which show the intermediate properties of metals and non-metals.
Example-Silicon, Boron
COMPOUND-Combination of two or more elements in a fixed proportion.
The new substance has totally different properties.
NCERT SOLUTIONS
Page: 15
1. What is meant by a substance?
Answer-
A pure single form of matter is called Substances. A substance has definite properties and compositions. Example – water.
2. List the points of differences between homogeneous and heterogeneous mixtures.
Answer-
Difference between homogeneous and heterogeneous mixture are :
Homogeneous mixture | Heterogeneous mixture |
Particles are uniformly distributed. | Non-uniform distribution of particles. |
Separation of particles by distillation method. | Separation of particles by filtration method |
It has a Uniform composition
They are not visible by naked eyes but we can see it by microscope. Example-sugar/salt water mixture. |
Non uniform composition.
Visible by naked eyes. Example- sand -water mixture. |
Page: 18
1. Differentiate between homogenous and heterogeneous mixtures with examples.
Answer-
Homogeneous mixture | Heterogeneous mixture |
Particles are uniformly distributed. | Non-uniform distribution of particles. |
Separation of particles by distillation method. | Separation of particles by filtration method |
It has a Uniform composition.
They are not visible by naked eyes but we can see it by microscope. Example-sugar/salt water mixture. |
Non uniform composition. Visible by naked eyes. Example- sand -water mixture. |
2. How are sol, solution and suspension different from each other?
Answer-
The key difference between sol, solution and suspension are:
SOL-
• They are heterogeneous in nature.
•Particle size are between 10 ^- 7 to 10 ^- 5 cm.
• They are visible and scatter light (Tyndall effect)
•Stable.
SOLUTION-
• Homogeneous in nature
• Less than particle size is less than 1 nm.
• Non-visible in nature and do not scatter light.
• Stable.
SUSPENSION
• Heterogeneous in nature.
• Particle size is more than 100 nm.
• Visible by naked eyes and scatters beam of light.
• Unstable.
3. To make a saturated solution, 36g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Ans:
Mass of sodium chloride (solute) = 36 g
Mass of water( solvent ) = 100 g
Mass of solution = mass of sodium chloride +mass of water
=100g+36g
=136 g
Example- sugar/salt water mixture Example- sand -water mixture.
Concentration is given by = Mass of solute/Mass of solution x 100
Concentration = 36/136 x 100
= 26.47%
Therefore, concentration of the solution is 26.47%
Page: 24
1. How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?
Answer:
Method for separation of a mixture of kerosene and petrol (miscible with each other)is distillation.
2. Name the techniques used to separate the following:
(a) Butter from curd.
(b) Salt from seawater
(c) Camphor from salt
Answer:
(a) Centrifugation.
(b) Evaporation
(c) Sublimation
3-What type of mixtures are separated by the technique of crystallization?
Answer:
They separate a pure solid in the form of its crystals from a solution.
Page 24
1. Classify the following as physical or chemical changes:
- Cutting of trees
- Melting of butter in a pan
- Rusting of almirah
- Boiling of water to form steam
- Passing of electric current through water and water breaking into hydrogen and oxygen gases.
- Dissolving common salt in water
- Making a fruit salad with raw fruits, and
- Burning of paper and wood
Answer:
•Cutting of trees-Physical change.
• Melting of butter in a pan-Physical change.
• Rusting of almirah-chemical change
• Boiling of water to form steam-Physical change
• Passing of electric current through water and water breaking into hydrogen and oxygen gases.-chemical change
• Dissolving common salt in water-physical change
• Making a fruit salad with raw fruits-chemical change
• Burning of paper and wood-chemical change
2. Try segregating the things around you as pure substances and mixtures.
Answer:
Pure substance- Salt, Water, Gold, Iron
Element-Brass, Milk, Saltwater, Soil
Exercise
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.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.
Answer:
(a) Evaporation
(b) Sublimation
(c) Filtration
(d) Chromatography
(e) Centrifugation
(f) Separating funnel
(g) Filtration
(h) Magnetic separation
(i) Winnowing/ sedimentation
(j) Decantation and filtration
2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate, and residue.
Answer:
Water-solvent
Tea leaves,sugar-solute
•Take a vessel and add a glass of water(solvent,) and put it in gas.
•Add tea leaves(solute) to the water and continue to heat. Tea leaves are insoluble on it.
•Now add sugar in it which is also solute. After a few seconds, it dissolved in water.
•At last add a cup of milk in it which also dissolve in water.
•Now tea is ready to serve. Filter tea in a cup with the help of strain and tea leaves
remains in residue as they are insoluble.
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 a 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 | 21 | 32 | 62 | 106 | 167 |
Sodium chloride | 36 | 36 | 36 | 37 | 37 |
Potassium chloride | 35 | 35 | 40 | 46 | 54 |
Ammonium chloride | 24 | 37 | 41 | 55 | 66 |
(a) What mass of potassium nitrate would be needed to produce a saturated solution of
potassium nitrate in 50 grams of water at 313K?
(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.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this
temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Answer:
(a) Mass of potassium nitrate need to produce saturated solution=?
Required amount= 62 x 50/100
= 31g of potassium nitrate.
(b) Pragya observed that the crystals of potassium chloride which would have compressed its solubility at low temperatures.
(c) Solubility of Potassium nitrate = 32/100=.32
The solubility of Sodium chloride =36/10=.36
The solubility of Potassium chloride = 35/100=.35
The solubility of Ammonium chloride = 37/100=.37
The highest amount of solubility= ammonium chloride
(d) The solubility of salt increases with increase in temperature.
4. Explain the following giving examples.
(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) suspension
Answer:
(a) Saturated solution-solution in which no more solute can be dissolved in solvent.
(b) Pure substance: Substances are a pure single form of matter. A substance has definite properties and compositions.
(c) Colloid: Solution in which solute particles have a size smaller than that of the suspension but larger than that of a solution. We cannot see it by naked eyes. But, these particles can easily scatter a beam of visible light.
(d) Suspension: It is a heterogeneous mixture in which solute particles do not dissolve and they remain suspended throughout the medium. We can see these particles with naked eyes.
5. Classify each of the following as a homogeneous or heterogeneous mixture.
soda water, wood, air, soil, vinegar, filtered tea.
Answer:
Homogeneous mixture-soda water, vinegar, filtered tea, air
Heterogeneous mixture-wood, soil.
6-How would you confirm that a colourless liquid given to you is pure water?
Answer:
If the given colourless liquid boils at 100 degree Celsius than it is pure water If the temperature deviates from 100-degree Celsius than it is not pure.
7. Which of the following materials fall into the category of “pure
substance”?
(a)Ice
(b)Milk
(c)Iron
(d)Hydrochloric acid
(e)Calcium oxide
(f)Mercury
(g)Brick
(e)Wood
(f)Air.
Answer:
Iron, Ice, Hydrochloric acid, Calcium oxide, Mercury are pure substances.
8. Identify the solutions among the following mixtures.
(a) Soil-
(b) Sea water
(c) Air
(d) Coal
(e) Soda water
Answer:
Sea water, air and soda water are mixtures.
9. Which of the following will show the “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.
Answer:
Starch solution and milk.
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:
(a) Element
(b) mixture
(c) mixture
(d) elements
(e) compounds
(f) elements
(g) elements
(h) mixture
(i) mixture
(j) mixture
(k) compound
(l) compounds
(m) mixture