Is Matter Around Us Pure? | NCERT Class 9 Science

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In-Text Questions & Answers - NCERT - Is Matter Around Us Pure?

Exercises Questions & Answers - Is Matter Around Us Pure?

Is Matter Around Us Pure? | NCERT Class 9 Science Chapter 2 | In-Text Questions & Answers | pdf | Exercises Questions & Answers

Q1. How do you classify matter around us?

Matter in our surroundings is divided into three groups. These are 1. element, 2. compound, & 3. mixture. These groups are further divided into different sub-groups. Details are depicted in the following picture –

Q2. What is element?

If no other matter (i.e. matter having different properties) is obtained after thorough analysing of a particular matter, then that matter is called an element. For example – Hydrogen, Oxygen, Nitrogen, Silver, Gold, Arsenic, Sodium, Iron, Calcium, Magnesium etc. There are more than 100 elements in the world.

Q3. What is inert element?

Some elements never take part in chemical reaction. These elements are called inert elements. For example – Helium, Neon, Argon, Krypton, Xenon and Radon. They are also called noble gas.

Q4. How do you classify element?
Element is divided into three categories ⇒ i) Metal, ii) Non-Metal, iii) Metalloid.

Q5. What is metal?

Copper, Silver, Gold, Mercury, Sodium, Aluminium, Iron etc. are examples of metal. Elements belonging to metal group are generally solid and heavier than non-metals.

Q6. What are the properties of metal?

Elements are solid and shiny. It also reflects light.
Melting and boiling point of metal are high.
Metal is hard, heavy and expandable. It can be hammered into sheet.
Metal is a good conductor for heat and electricity.
Metal produces sound if hammered.
During electrolysis process metal produces (+) positive ion and accumulates at cathode.
Oxide produced by metal and oxygen is either basic or acidic or both.

Exceptions-

Mercury is liquid like non-metal.
Melting point of Caesium and Gallium is very less (30^OC).
Sodium, Potassium, Lithium etc. are soft and lighter in weight.

Q7. What is non-metal?

Oxygen, Nitrogen, Hydrogen, Carbon, Sulphur, Bromine, Chlorine, Phosphorus etc. are examples of non-metal.

Q8. What are the properties of non-metal?

It is liquid or gas and is not shiny. It also can’t reflect light.
Melting and boiling point are lower.
Non-metal is lightweight, soft and breakable. It also can’t be hammered into sheet.
Non-metals are poor conductor for heat and electricity.
It does not produce sound if hammered.
In case of electrolysis, non-metal produces (-) negative ion and accumulates at anode.
Oxide produced by non-metal and oxygen is either acidic or neutral.

Exceptions-

Bromine is normally liquid.
Carbon, Sulphur, Phosphorus and Iodine are normally solid.
Melting point of non-metals like Carbon, Boron, Silicon etc. are high like metal.

Q9. What is metalloid?

Metal which has properties of both metal and non-metal is called metalloid. For example – Tin, Arsenic, Antimony etc.

Q10. What is compound and it’s classification?

Two or more element produce a compound after taking part in chemical reaction in a certain proportion. Properties of compound differ from the individual properties of it’s constituents. For example – Ammonia [NH₃⇒ Nitrogen(N) : Hydrogen(H) = 14:3], Water [H₂O ⇒ Hydrogen (H) : Oxygen(O)=1:8] etc.

Q11. How do you classify compound?

Compound is of two types -i) Inorganic Compound, & ii) Organic Compound.

Inorganic Compound is related to chemical reaction of two or multiple elements. For example – Sulphuric Acid, Nitric Acid, Sulphur dioxide, Water, Sodium Chloride, Carbon dioxide, Ammonia etc.
Organic Compound are related to plants and animals. It is noted that every organic compound has carbon and hydrogen as it’s common constituents. For example – Acetic acid, Methane, Petroleum, Sugar, Coal etc.

Q12. What are the properties of compound?

Boiling or melting point of a compound is fixed.
Compound is homogeneous.
Constituents lose their own properties during formation of compound. That’s why compound has totally different properties from it’s constituents.
The constituents of a compound always take part in chemical reaction in a certain ratio to produce that compound. Alteration in ratio of constituents is not possible.
Production of compound in chemical reaction is always either exothermic or endothermic.

Two or multiple element or compound mix with each other at any proportion and produces a mixture. We divide mixture into two types ⇒ 1) Heterogeneous Mixture, & 2) Homogeneous Mixture.

Q13. What is mixture and how do you classify compound?

Mixture in which proportion of it’s constituents is not same or uniform in every parts or portion of it is called heterogeneous mixture. For example – Sand and sugar, Iron and Sulphur, Smoke (Air + various gases + particles) etc.

Mixture in which proportion of it’s constituents is equal & uniform in every parts or portions of it is called homogeneous mixture. For example – Air (Oxygen + Nitrogen + Carbon dioxide + other gases), Lemonade (Sugar + Carbon dioxide + Water), Brass (Copper + Zinc), Bronze (Copper + Tin) etc.

Q14. What are the properties of mixture?

It has no fixed boiling or melting point.
Production of mixture is neither exothermic nor endothermic.
Constituents of mixture mix with each other at any proportion (except solution).
Properties of constituents of a mixture remain unaltered.
Constituents of mixture can be separated from mixture through physical process such as sedimentation and decantation, sublimation, evaporation, distillation, filtration etc.

Q15. What is solution?

By mixing with each other, two or more matter produce a homogeneous substance which is called solution. Every part of a solution contains it’s constituents in same proportion. Solution has two parts – i) Solute, & ii) Solvent. Quantity of solute is less and quantity of solvent is more in a solution. For example – Salt (solute) + Water (solvent), Sugar ( solute ) + Water ( solvent ) etc.

Solute + Solvent = Solution

Q16. Write down properties of solution.

It is homogeneous. Every parts of a solution has same proportion of it’s constituents.
Production of solution may or may not be exothermic or endothermic.
Properties of solute and solvent remain same in case of a solution.
You can change quantity of solute in a solution by changing the temperature of solution.
Diameter of solute particle is of range of 10^-7 mm.

Q17. What are the examples of different type of solutions?

Solid solute + Solid solvent ⇒ In this case, two or multiple metal melt and solidify to make a homogeneous mixture. For example – Brass (Copper + Zinc), Bronze (Copper + Tin) etc.

Solid solute + Liquid solvent ⇒ Solid mixes with liquid and produces homogeneous mixture or solution. For example – Salt + Water, Sugar + Water etc.

Liquid solute + Liquid solvent ⇒ Liquid mixes with liquid and produces homogeneous mixture or solution. For example – glycerin solution, Diluted acid etc.

Gas solute + Solid solvent ⇒ Metal like Nickel, Palladium etc. absorbs hydrogen gas in heated condition and produces homogeneous mixture.

Gas solute + Liquid solvent ⇒ Chlorine, Carbon dioxide, Sulphur dioxide etc. mix with water.

Liquid solute + Solid solvent ⇒ Metal like Sodium, Gold, Silver, Zinc etc. produces amalgam mixture with Mercury.

Gas solute + Gas solvent ⇒ Two or more gases mix with each other without chemical reaction and produce homogeneous mixture. For example – Air.

Q18. How do you classify solution?

Solution is classified into three categories – A) unsaturated solution, B) saturated solution, & C) supersaturated solution.

Q19. What is unsaturated solution?

Unsaturated solution is the solution in which you can add more solute at a certain temperature. If the proportion of solute is less, then it is dilute solution. For example – dilute acid. If proportion of solute is more, then it is concentrated solution. For example – concentrated acid.

Q20. What is saturated solution?

Saturated solution is the solution in which you can not add any more solute at certain temperature.

Q21. What is supersaturated solution?

The solution in which quantity of solute is more than it’s saturated level at a certain temperature is the supersaturated solution.

Q22. What is solubility?

At a certain temperature, maximum quantity of solute which can be added to 100 gram of solvent is called solubility of that solute at said temperature. In other words solubility is the actual quantity of solute in a saturated solution at a certain temperature.

What does ‘solubility of salt is 35 at 20^OC’ mean? ⇒ It means that you can add maximum 35 gram of salt in 100 gram water at 20^OC. If you add salt further into water, that added salt will not mix with water and will be in solid state.

Q23. What is concentration and how do you express concentration in percentage (%)?

Concentration is the percentage of solute present in a solution. We express concentration in percentage(%) in following three ways –

A) [Mass/mass] concentration when both solute and solution are expressed in mass.

B) [Mass/volume] concentration when solute is expressed in mass and solution is expressed in volume.

C) [Volume/volume] concentration when both solute and solution are expressed in volume.

Example: In a solution, there are 125 gram acid and 500 gram water at 25^OC. Find the solubility and concentration.
Answer: Given that ⇒ solute = 125 gram, solvent = 500 gram, Sslution = (100 + 500) gram = 600 gram.
So ⇒ solubility = (125/500) X 100 = 25, & concentration = (125/600) X 100 % = 20.83%.

Q24. How do temperature and pressure affect solubility?

Generally solubility of solids increases with the rise in temperature and decrease with the fall in temperature. For example – solubility of Potassium Nitrate, Copper Sulphate, Ammonium Chloride in water increases with rise in temperature.

# Generally, solubility of solid and liquid in liquid do not change significantly with the rise or fall in pressure.
# On the other hand, solubility of gas in liquid increases with the rise in pressure and decreases with the fall in pressure.

Though there are some exceptions such as –
a) In general solubility remains almost same with the rise or fall in the temperature of salt and Potassium dichromate solution.
b) Solubility of Calcium Sulphate in water decreases with the rise in temperature.
c) At first solubility of Glauber salt in water increases up to 32^OC. It decreases after 32^OC due to chemical change.

Q25. How do you identify a unsaturated or saturated solution?

Add small quantity of solute in a solution. If the solute melts and mixes in solution homogeneously, then it is unsaturated solution. But if solute is not melted and remains in it’s original state, then it is saturated solution. Take a glass of water and gradually add salt in small quantity. Salt will be melted until the solution becomes saturated. After some time you will notice that melting of salt is stopped. Therefore, we say that solution has become saturated.

Q26. What is colloidal solution? What are the properties of colloids?

Colloidal solution is heterogeneous mixture where diameter of solute particle ranges from 10^-4mm to 10^-6mm. For example – milk, jelly, fog, foam, sponge etc. We can not see particles through naked eye.

Properties of colloids ⇒
1. It is heterogeneous mixture.
2. It is not possible to see the particles in a colloidal solution.
3. It scatters a beam of light passing through it & makes its path visible.
4. Colloids are stable because the solute particles do not settle down when left undisturbed.
5. It can not be separated from the mixture through filtration process. We need a special technique called centrifugation for this purpose.

Q27. What is suspension? What are the properties of suspension?

Answer: Suspension is a heterogeneous mixture where solute particle does not dissolve. In that case we can see particles through naked eye.

Properties of suspension ⇒
1. It is heterogeneous mixture.
2. The solute particle does not dissolve & remain suspended throughout the bulk of the medium.
3. The solute particles scatter of a suspension scatter a beam of light passing through it & makes its path visible.
4. Suspension is unstable because the solute particles settle down when left undisturbed.
5. A suspension breaks when he solute particles settle down. After that it cannot scatter light any more.
6. Filtration is the process to separate solute particles from a suspension.

Q28. What is alloy?
Alloy is a mixture of metal with metal or metal with non-metal. We can not separate the constituents of an alloy by physical process. It shows the properties of its constituents. It can also have variable composition of its constituents. For example – brass which has 30% zinc & 70% copper as constituents.

Q29. What is Tyndall effect?

The particles in a colloidal solution can easily scatter a beam of light. This light scattering by particles in a colloid is called Tyndall effect. Tyndall effect is named after 19th-century Irish physicist John Tyndall who discovered it. We can observe Tyndall effect when a fine beam of light enters a room through a small hole. In this case, the particles of dust & smoke in the air cause the scattering of light. We can also observe Tyndall effect when sunlight passes through the canopy of a dense forest. In this case, the tiny droplets of mist cause the scattering of light.

Q30. What are the components of a colloidal solution? Write down examples of different types of colloids.

A colloidal solution has two components. These are – a. the dispersed phase, & b. the dispersing medium. The dispersed phase is formed by the dispersed particles i.e. solute-like component. The dispersing medium is the other component in which the dispersed phase is suspended. We classify colloids as per the state (i.e. solid, liquid, gas) of the dispersing medium & the dispersed phase. Following are some examples –

Dispersed Phase	Dispersed Medium	Type of Colloid	Examples
Solid	Solid	Solid-Sol	Coloured gemstone, milky glass etc.
Solid	Liquid	Sol	Mud, Milk of magnesia etc.
Solid	Gas	Aerosol	Automobile exhaust, smoke etc.
Liquid	Solid	Gel	Butter, jelly, cheese etc.
Liquid	Liquid	Emulsion	Face cream, milk etc.
Liquid	Gas	Aerosol	Mist, clouds, fog etc.
Gas	Solid	Foam	Foam, sponge, rubber, pumice etc.
Gas	Liquid	Foam	Shaving foam etc.

Q31. How to separate components of air?
Air is a homogeneous mixture. We can separate its components through fractional distillation. At first air is compressed by increasing pressure & cooled by decreasing temperature. Thus liquid air is formed. Then that liquid air is warmed-up slowly in a fractional distillation column. As a result, we get different gases of air at different heights depending upon their boiling temperature. For example, we obtain oxygen at -183^oC as its boiling point is -183^oC. Similarly argon & nitrogen are obtained at -186^oC & -196^oC respectively.

Q32. What is crystallization?

Crystallization is a method to separate a pure solid in the form of crystals from a solution. This process is used to remove impurities from sea salt & to separate crystals of alum (phitkari) from impure samples. This process is chosen according to the nature of a mixture.

Q33. Name some techniques to separate the components of a mixture.

Some techniques are –

Evaporation is a process to separate volatile component (solvent) from its non-volatile solute.

Centrifugation is a process where denser particles are forced to the bottom & lighter particles stay at the top.

Chromatography is a process for separation of those solutes that dissolve in same solvent.

Distillation is the technique used to separate components of a mixture containing two miscible liquids that boil without decomposition having sufficient temperature difference in their boiling points.

Q34. What is the flowchart for classification of matter?

In-Text Questions & Answers | NCERT Class 9 Science Chapter 2 | Is matter around us pure?

In-Text Questions & Answers | NCERT Class 9 Science Chapter 2 | Is matter around us pure? | Page No. 15

1) What is meant by a substance?
Pure substance is a material that consists of a single type of particles. These particles of a pure substance have identical chemical properties.

2) List the points of differences between homogeneous & heterogeneous mixtures.
Write down Q13.

In-Text Questions & Answers | NCERT Class 9 Science Chapter 2 | Is matter around us pure? | Page No. 18

1. Differentiate between homogeneous & heterogeneous mixtures with example.
Read Q13 in detail.

2. How are sol, solution & suspension different from each other?

Answer:	Sol	Solution	Suspension
A)	Heterogeneous in nature.	Homogeneous in nature.	Heterogeneous in nature.
B)	Shows Tyndall effect & scatters a beam of light.	Does not show Tyndall effect & does not scatter a beam of light.	Does not show Tyndall effect & does not scatter a beam of light.
C)	It is impossible to see particles with naked eyes.	Solution dissolves particles.	It is possible to see particles with naked eyes.
D)	Quite stable.	Very stable.	Unstable.
E)	For example - milk, blood etc.	Example - sugar in water, salt in water etc.	For example - sand in water etc.

3) To make a saturated solution, 36 gram of sodium chloride is dissolved in 100 gram of water at 293 K. Find it’s concentration at this temperature.
Read Q23 in detail. In this case ⇒ at 293 K ⇒ solute = sodium chloride = 36 gram, solvent = water = 100 gram, solution = (36+100) gram = 136 gram.
So concentration = [(solute/solution) x 100] % = [(36/136) x 100] % = 26.47 %.

In-Text Questions & Answers | NCERT Class 9 Science Chapter 2 | Is matter around us pure? | Page No. 24

1) How will you separate a mixture containing kerosene & petrol (difference in their boiling points is more than 25 ⁰C, which are miscible with each other?

Distillation is the process to separate a mixture containing kerosene & petrol. There is 25 ⁰C difference in boiling points of kerosene & petrol. Kerosene has lower boiling temperature than petrol.

In this case we need a distillation flask with thermometer, a beaker, a water condenser & a Bunsen burner. The mixture is kept in the distillation flask. Then we start heating the distillation flask. As a result kerosene starts vapourising. It is then passed through water condenser & collected in beaker. At the end only petrol resides in the distillation flask.

2) Name the technique to separate – a) butter from curd, b) salt from sea-water, c) camphor from salt.

a) Butter from curd :- We can separate butter from through centrifugation process. It is the process in which denser particles are forced to the bottom & the lighter particles remain at the top when spun fast.
b) Salt from sea-water :- Evaporation is the process to obtain salt from sea water. Due to evaporation, the quantity of water becomes less to keep salt particles apart. This causes re-crystallisation of the salt.
c) Camphor from salt :- Sublimation is the process to obtain camphor from salt. In this case, heating is done. As a result, camphor turns into gaseous state from solid state & it is collected.

3) What type of mixtures are separated by the technique of crystallisation?
Basically it is a process to separate a pure solid in the form of it’s crystals from a solution. For example – obtaining salt from sea water, obtaining Phitkari or alum from impure samples etc.

In-Text Questions & Answers | NCERT Class 9 Science Is matter around us pure? | Page No. 24

1) Classify the following as chemical or physical changes:
a) cutting of trees,
b) melting of butter in a pan,
c) rusting of almirah,
d) boiling water to form steam,
e) passing of electric current, through water & the water breaking down into hydrogen & oxygen gases,
f) dissolving common salt in water,
g) making a fruit salad with raw fruits,
h) burning of paper & wood.
At first we should know the difference between chemical change & physical change. Chemical change or chemical reaction changes the properties of the matters & forms new substances. On the other hand, physical change is only change in states of matter without any change in composition & chemical properties.

Chemical changes ⇒
c) rusting of almirah,
e) passing of electric current, through water & the water breaking down into hydrogen & oxygen gases,
h) burning of paper & wood.

Physical change ⇒
a) cutting of trees,
b) melting of butter in a pan,
d) boiling water to form steam,
f) dissolving common salt in water,
g) making a fruit salad with raw fruits.

2) Try segregating the things around you as pure substances or mixtures.
Generally we buy daily necessities from nearby grocery regularly. You may easily identify what is pure by checking ingredients written on the packet. For example – salt, sugar, water etc. are pure substance. On the other hand milk, butter, air, cold drink, lemon water, rubber, food etc. are impure & mixtures.

Exercises Questions & Answers | NCERT Class 9 Science Chapter 2 | Is matter around us pure? | Page Nos. 28 - 30

1. Which separation technique 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 & 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 pin from sand.
i) Wheat grains from husk.
j) Fine mud particles suspended in water.

a) Sodium chloride from its solution in water ⇒ Evaporation or Crystallisation.
b) Ammonium chloride from a mixture containing sodium chloride & ammonium chloride ⇒ Sublimation.
c) Small pieces of metal in the engine oil of a car ⇒ Centrifugation or Sedimentation.
d) Different pigments from an extract of flower petals ⇒ Chromatography.
e) Butter from curd ⇒ Centrifugation.
f) Oil from water ⇒ Separating funnel.
g) Tea leaves from tea ⇒ Manual or hand picking.
h) Iron pin from sand ⇒ Using magnet.
i) Wheat grains from husk ⇒ Winnowing.
j) Fine mud particles suspended in water ⇒ Centrifugation.

2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate & residue.

Answer: Put the solvent (i.e. water) in a kettle & start heating. At the time the solvent boils, add some solute i.e. milk. Thus a solution is prepared. Then pour some tea leaves over a sieve & start pouring the hot solution over it. Colour of tea leaves goes into solution as filtrate & remaining tea leaves remain as insoluble residue. After that add some sugar into the solution. The sugar easily dissolves into the solution & the solution is ready to serve.

3. Pragya tested the solubility of three different substances at different temperatures & collected the as given in the following picture - ( results in table shows quantity in grams dissolved in 100 grams of water to form a saturated solution) -

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: At 313 K, to make a saturated solution 100 gram water requires 62 gram potassium nitrate. So at 313 K, 50 gram water requires [62 x (50/100)] gram = 31 gram potassium nitrate.

b) Pragya makes a saturated solution of potassium chloride in water at 353 K & leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

Answer: Observe row of potassium chloride thoroughly. Solubility of potassium chloride at 353 K is 54. But it reduces when temperature gets reduced (35 at 283 K). So if the saturated solution of potassium chloride at 353 K is cooled down & brought back to room temperature (around 300 K), then some amount of potassium chloride reappears as undissolved solid.

c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

Answer: Specifically read column 3 . So the answer is – a) Potassium nitrate – 32, b) Sodium chloride – 36, c) Potassium chloride – 35, & d) Ammonium chloride – 37. Ammonium chloride has highest solubility at 293 K i.e. 37.

d) What is the effect of change of temperature on the solubility of a salt?

Answer: Solubility increases with the increase in temperature & decreases with the decrease in temperature.

4. Explain - a) saturated solution, b) pure substance, c) colloid, d) suspension.

Answer: a) Saturated solution – Q20 & Q25.
b) Pure substance – It is a material that consists of a single type of particles. These particles of a pure substance have identical chemical properties.
c) Colloid – Q26.
d) Suspension – Q27.

5. Classify each as a homogeneous or heterogeneous mixture - soda water, wood, air, soil, vinegar, filtered tea.

Answer: Homogeneous mixture – soda water, air, vinegar, filtered tea. Heterogeneous mixture – wood, soil.

6. How would you confirm that a colourless liquid given to you is pure water?

Answer: Boiling point of water is 100 ⁰C. Let start heating the colourless liquid. If it boils at 100 ⁰C, then it is pure. If it boils at lower or higher than 100 ⁰C, then it is impure or added by some impurities.

7. Which are pure substance - a) Ice, b) Milk, c)Iron, d) Hydrochloric acid, e) Calcium oxide, f) Mercury, g) Brick, h) Wood, i) Air?

Answer: Pure substances are – a) Ice, c)Iron, d) Hydrochloric acid, e) Calcium oxide, & f) Mercury.

8. Identify the solutions among the mixtures - a) soil, b) Sea water, c) Air, d) Coal, e) Soda water?

Answer: Solutions are b) Sea water, c) Air, & e) Soda water.

9. Which shows 'Tyndall effect' - a) Salt solution, b) Milk, c) Copper sulphate solution, d) Starch solution?

Answer: Tyndall effect is shown by – b) Milk, & d) Starch solution.

10. Classify into elements, compounds & 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: Elements – a) Sodium, d) Silver, f) Tin, & g) Silicon.
Compounds – e) Calcium carbonate, k) Methane, 7 l) Carbon dioxide.
Mixtures – b) Soil, c) Sugar solution, h) Coal, i) Air, j) Soap, & m) Blood.

11. Which are chemical changes - a) Growth of a plant, b) Rusting of iron, c) Mixing of iron fillings & sand, d) Cooking of food, e) Digestion of food, f) freezing of water, g) Burning of a candle?

Answer: Chemical changes – a) Growth of a plant, b) Rusting of iron, d) Cooking of food, e) Digestion of food, & g) Burning of a candle.

# Is matter around us pure? ⇒ Extra mathematical questions ⇒

1. 8 gram substance is dissolved in 40 gram water at 50⁰C, to prepare a saturated solution. Find the solubility of that substance at 50⁰C.

Answer: Firstly, go through Q22. Solubility of the substance at 50⁰C = [(8/40) x 100] = 20.

2. 90 gram saturated water solution (at 45⁰C ) gives 30 gram substance in boiling. Find solubility of that substance at 45⁰C.

Answer: During boiling, water vapourises and 30 gram solute is obtained. So quantity of water was = (90-30) = 60 gram. So solubility of the substance at 45⁰C = [(30/60)x100 ] = 50.

3. Solubility of salt is 40 at 60⁰C. At same temperature how much salt is required to prepare 560 gram solution? Find the concentration also.

Answer: Solubility of salt is 40 at 60⁰C means that 40 gram salt is added into 100 gram water to prepare saturated solution. So, total mass of the solution is ( 40 + 100 ) i.e. 140 gram.
Therefore, 140 gram saturated solution needs 40 gram salt
⇒ 1 gram saturated solution needs (40/140) gram salt
⇒ 560 gram saturated solution needs 560 x (40/140) = 160 gram salt.

Hence, Concentration = (Mass of solute/ Mass of solution) x 100% = (140/560) x 100 % = 25 %

Is Matter Around Us Pure? – To Be Continued – Is Matter Around Us Pure?