NCERT Exemplar Solution Class 9 Science Chapter 2
NCERT Exemplar Solution Class 9 Science Chapter 2 Is Matter Around Us Pure all Questions Answer Solution. NCERT Exemplar Class 9 Science Chapter 2 Solution PDF.
NCERT Exemplar Solution Class 9 Science Chapter 2: Overview
NCERT Exemplar Solution Class 9 Science Chapter 2 |
|
Board |
NCERT |
Topic |
Exemplar Problem Solution |
Class |
9 |
Subject |
Science |
Chapter |
2 |
Chapter Name |
Is Matter Around Us Pure |
NCERT Exemplar Solution Class 9 Science Chapter 2 Is Matter Around Us Pure
Multiple Choice Questions
1) Which of the following statements are true for pure substances?
i) Pure substances contain only one kind of particles
ii) Pure substances may be compounds or mixtures
iii) Pure substances have the same composition throughout
iv) Pure substances can be exemplified by all elements other than nickel.
a) I) and II)
b) I) and III)
c) III) and IV)
d) II) and III)
Answer:-b) I and III
The true statements about the pure substances are pure substances contain only one kind of particles and pure substances have the same composition throughout.
2) Rusting of an article made up of iron is called
a) Corrosion and it is a physicalas well as chemical change
b) Dissolution and it is a physical change
c) Corrosion and it is a chemical change
d) Dissolution and it is a chemical change
Answer:- c) corrosion and it is a chemical change.
Corrosion of an iron is a known as Rusting and it is a chemical change, it is carried out due to reaction of iron with water and oxygen . After reaction it shows change in colour, also the product of these reaction is hydrate iron oxide which is different from the iron element.
3) A mixture of sulphur and carbon disulphide is
a) Heterogeneous and shows Tyndall effect
b) Homogeneous and shows Tyndall effect
c) Heterogeneous and does not show Tyndall effect
d) Homogeneous and does not show Tyndall effect
Answer:- d) homogeneous and does not show Tyndall effect
A mixture of sulphur and carbon disulphide is homogeneous mixture. Homogeneous mixture is not an colloidal Solution therefore due to very small particle size light does not scatter through the solution and thus it does not show Tyndall effect.
4) Tincture of iodine has antiseptic properties. This solution is made by dissolving
a) Iodine in potassium iodide
b) Iodine in Vaseline
c) Iodine in water
d) Iodine in alcohol
Answer:- d) Iodine in alcohol
Iodine does not dissolve in water, and Vaseline. Tincture iodine is made by mixing iodine into alcohol, as alcohol is an good solvent . Thus this solution is made by dissolving iodine into alcohol.
5) Which of the following are homogenousin nature?
i) Ice
ii) Wood
iii) Soil
iv) Air
a) I and III
b) II and IV
c) I and IV
d) III and IV
Answer:- c) I and IV
Ice and Air is homogeneous in nature. While the other two wood and soil are not homogeneous. Air and Ice is homogeneous because the particles size is so small that cannot visible by naked eyes, also it cannot differentiate from one another.
6) Which of the following are physical changes?
i) Melting of iron metal
ii) Rusting of iron
iii) Bending of an iron rod
iv) Drawing a wire of iron metal
a) I, II, and III
b) I, II, and IV
c) I, III and IV
d) II, III, and IV
Answer:- c) I, III, and IV
Melting of iron metal, bending of an iron rod and Drawing a wire of iron metal are physical changes because the changes does not contain any chemical reaction or chemical change after the process. Only the Rusting of iron is an chemical change because rusting effect of an chemical reaction.
7) Which of the following are chemical changes?
i) Decaying of wood
ii) Burning of wood
iii) Sawing of wood
iv) Hammering of a nail into a piece of wood
a) I and II
b) II and III
c) III and IV
d) I and IV
Answer:- a) I) and II
Sawing of wood and hammering of a nail into a piece of wood is an physical changes. Decaying of wood and burning of wood are chemical changes because these occursafter an chemical reaction in it.
8) Two substances, A and B were made to react to form a third substance, A2B according to the following reaction
2A + B → A2 B
Which of the following statements concerning this reaction are incorrect?
i) The product A2 B shows the properties of substances A and B.
ii) The product will always have a fixed composition.
iii) The product so formed cannot be classified as a compound.
iv) The product so formed is an element.
a) I, II, and III
b) II, III, and IV
c) I, III, and IV
d) II, III and IV
Answer:- c) I, III and IV
Only the option II product will always have a fixed composition is correct among the above. Other all the options are incorrect. In the reaction 2 molecules of A and 1 molecule of B formed the product A2B which is in fixed ratio.
9) Two chemical species X and Y combine together to form a product P which contains both X and Y.
X + Y → P
X and Y cannot be broken down into simpler substances by simple chemical reactions. Which of the following concerning the species X, Y and P are correct?
i) P is a compound.
ii) X and Y are compounds
iii) X and Y are elements
iv) P has a fixed composition
a) I, II,and III
b) I, II, and IV
c) II, III, and IV
d) I, III, and IV
Answer:- d) I, II, and III
I, II and III are correct regarding the X,Y and p. as per given information p can be broken into elements thus it is formed by combining elements hence p is an compound. And X and Y cannot broken down into pieces thus these are elements.
Short Answer Questions
10) Suggest separation technique(s) one would need a employ to separate the following mixtures.
a) Mercury and water
Answer:-
Mercury and water can be separated by the process of decantation using a separating funnel. Mercury is heavier than water and therefore it settle down to the bottom of the separating funnel and forms a separate layer. Thus it can be easily separated.
Process of decantation
b) Potassium chloride and ammonium chloride
Answer:-
Potassium chloride and ammonium chloride can be separated by using the method of Sublimation. Ammonium chloride converted directly from solid to gaseous state after rising the temperature and potassium chloride remains as it is. Thus the gaseous ammonium chloride can be collected in the inverted funnel and potassium chloride remains in the beaker.
Process of Sublimation
c) Common salt, water and sand
Answer:- Common salt, water and sand can be separated by two processes perform led one after one. Common salt dissolves in water but sand does not.
Mixture of Common salt, sand and water
Here first filtration process is Carried out, so the sand remains on the filter paper and the filtrate i.e. salt and water taken out in beaker.
Process of filtration
Further the process of evaporation separate the water and the salt. Evaporation leads to the conversion of water into vapours and salt remains as it is.
Process of Evaporation
d) Kerosene oil, water and salt
Answer:- Kerosene oil, water and salt can be separated by using two methods. Salt dissolves in water and kerosene become lighter in weight it forms a separate layer over the water. First the process of decantation using separating funnel is used. By which the kerosene will be removed and the salt and water taken out in another beaker.
Process of decantation
Further the process of Evaporation carried out for the separation of salt and water. Water gets evaporated on the higher temperature and salt Remains as it is in the beaker.
Process of Evaporation
11) Which of the tubes in fig. 2.1 (a) and (b) will be more effective as a Condenser in the Distillation apparatus?
Answer:-
The tube A will be more effective condenser in the distillation apparatus. Because the tube A contains marbles which increases the surface area for cooling the vapours. Thus the vapours gets more time for condensation, when it comes in contact with marbles. Therefore the tube A is effective Condenser than tube B.
12) Salt can be recovered from its solution by evaporation. Suggest some other technique for the same ?
Answer:-
Salt is removed by the process of evaporation from the solution but there is also another method for removing salt is Crystallization. Crystallization is very effective method than the evaporation. Crystallization remove ls the impurities (Soluble) which does not removed by the evaporation process.
13) The ‘ sea -water ‘ can be classified as a homogeneous as well as heterogeneous mixture. Comment.
Answer:-
The upper layer of Sea water can be classified as homogeneous when we considered the salt and water only. Salt is totally soluble in the water and thus we can consider it as an homogenous mixture.
If we Classify the deep layers of sea water as an heterogeneous mixture then it is a Insoluble mixture of salt, water, decayed plants and sea animals, mud etc.
14) While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C) what technique can be employed to get back the acetone ? Justify your choice.
Answer:- For diluting solution of acetone the Distillation process can be carried out. Acetone is more volatile than the water as acetone has boiling point 56°C and water has 100°C. Then acetone will be evaporates first and collected to get the acetone back.
15) What would you observe when
a) A saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature.
Answer:- when a saturated solution of potassium Chloride prepared at 60°C is allowed to cool at room temperature then the solid potassium chloride will be separated from the solution as the temperature affect the solubility of the solid.
b) An aqueous sugar solution is heated to dryness.
Answer:-
When the aqueous solution is heated then first the water will be completely evaporated and then the sugar remained will get charred.
c) A mixture of iron filings and sulphur powder is heated strongly.
Answer:-
When a mixture of iron fillings and sulphur powder is heated strongly the iron and sulphide combines to form the iron sulphide.
16) Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do.
Answer:-
Colloidal solution has smaller size of particles than the suspension. Also, the interaction between the molecules in the suspension is not so strong it is relatively weak to keep the particles suspended around. And therefore it settles down to the bottom. On the other side the colloidal solution has strong molecular interactions therefore these particles remains suspended in the solution.
17) Smoke and fog both are aerosols. In what way are they different?
Answer:-
Smoke and fog both are gases and both are in dispersed medium i.e. continuous phase. But the difference between the smoke and fog is that, the dispersed phase in smoke is solid, while the dispersed phase in fog is liquid.
(18) Classify the following as physical or chemical properties.
(a) The composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements.
Answer:- it is a physicalproperty.
Steel is an alloy. Here the composition of an alloy is given, there is not any chemical reaction in it therefore it is a physical property.
(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
Answer:- it is a chemicalproperty.
Here zinc is dissolving in the hydrochloric acid and hydrogen gas is evolving on the completion of reaction thus the chemical change is observed. Therefore it is a chemical property.
(c) Metallic Sodium is soft enough to be cut with a knife.
Answer:-. It is a Physicalproperty.
There is no chemical reaction involving in the cutting with a knife therefore it is a physical property.
(d) Most metal oxides form alkalis on interacting with water.
Answer:- it is a chemical property.
Reaction between the metal oxides and alkalis are forming a new compound therefore it is an chemical property.
19) The teacher instructed three students ‘A’, ‘B’, and ‘C’ respectively to prepare a 50% (mass by volume) Solution of sodium hydroxide (NaOH). ‘A’ dissolved 50g of NaOH in 100 mL of water, ‘B’ dissolved 50g of NaOH in 100g of water while ‘C’ dissolved 50g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?
Answer:-
Student A and B have made 150 ml of solution but student C have make up the volume and made correct 100 ml of solution.
C has made the desired solution, and the conclusion is as below –
Mass bymass of solute
Volume %=–––––––––––––––– × 100
Volume of solution
50
= ––––––× 100
100
= 50% mass by volume
Therefore, student C has made the right 50% mass/Volume solution of NaOH.
20) Name the process associated with the following
a) Dry ice is kept at room temperature and at one atmospheric pressure.
Answer:- sublimation process
b) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
Answer:- Diffusion process
c) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
Answer:- diffusion process
d) A acetone bottle is left open and the bottle becomes empty.
Answer:-evaporation process
e) Milk is churned to separate cream from it.
Answer:- centrifugation process
f) Setting of sand when a mixture of sand and water is left undisturbed for some time.
Answer:- sedimentation process
g) Fine beam of light entering through a small hole in a dark room, illuminates the particles in its paths.
Answer:- the process of scattering of light (Tyndall effect).
21) You are given two samples of water labelled as ‘A’ and ‘B’. Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment.
Answer:- The water in sample B will not freeze at 0°C. Because it contains some impurities, thus it is not an pure water. Pure water has boiling point 100°C and freezing point 0°C at 1 atm. Pressure. But it sample B is not pure and therefore it will not freeze on 0°C. While the sample A which is a pure water will freeze at 0°C.
22) What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments?
Answer:-
24 carat pure gold is so soft, it can be bend easily. It does not have any strength. But the gold alloyed with silver or copper has the strength compared to pure gold. Adding copper or silver gives the strength and it does not bend easily.
23) An element is sonorous and highly ductile. Under which category would you classify this element? What other characteristics do you expect the element to possess?
Answer:-
An element which is highly ductile and sonorous can be classified under the category of metals. Metals also has other characteristics such as lustre, malleable, good Conductors of heat and electricity.
24) Give an example each of the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures.
a) A volatile and a non- volatile component.
Answer:- evaporation or distillation process can be used for the separation of a volatile and a non-volatile components.
b) Two volatile components with appreciable difference in boiling points
Answer:- distillation process is applied to separate two volatile components.
c) Two immiscible liquids
Answer:- the technique of separation by using separating funnel is used here.
d) One of the components changes directly form solid to gaseous state.
Answer:- Sublimation method is used toseparate the components which is directly changes from solid to gaseous state.
e) Two or more coloured constitutes soluble in some solvent.
Answer:- chromatographymethod is used for the separation of two or more coloured constitutes soluble in some solvent.
25) Fill in the blanks.
a) A colloid is a __________ mixture and its components can be separated by the technique known as ____________.
Answer:- A colloid is a heterogeneous mixture and it’s components can be separated by the technique known as centrifugation.
b) Ice, water and water vapour look different and display different ___________ properties but they are _________ the same.
Answer:- Ice, water and water vapour look different and display different physical properties but they are chemically the same.
c) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of ____________ and the lower layer will be that of ______________.
Answer:- A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of water and the lower layer will be that of chloroform.
d) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 k can be separating by the process called ________.
Answer:- A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25k can be separating by the process called fractional distillation.
e) when light is passed through the water containing a few drops of milk, it shows a bluish tinge. This is due to the __________ of light by milk and the phenomenon is called ___________. This indicates that milk is a ___________ solution.
Answer:- when light is passed through the water containing a few drops of milk, it shows a bluish tinge. This is due to the scattering of light by milk and the phenomenon is called Tyndall effect. This indicates that milk is a colloidal solution.
26) Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it be a pure substance or a mixture? Give reasons for the same.
Answer:- Yes, it will be a pure substance. Whether the sources are same or different the chemical composition of the sugar crystals is same, thus it is a pure substance.
27) Give some examples of Tyndall effect observed in your surroundings?
Answer:- The scattering of the beam of light is known as Tyndall effect. It can be seen when a beam of light passes through a heterogeneous mixture, it gets scattered.
Another example if Tyndall effect is that, when a light passes from a dark room or dense forest.
28) Can we separate alcohol dissolved in water by using a separating funnel?
If yes, then describe the procedure
If not, explain.
Answer:- No. We cannot separate the mixture of alcohol and water by separating funnel because the alcohol and water forms a miscible solvent. Thus it cannot be separated by separating funnel.
29) On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.
a) Is this a physical or a chemical change ?
Answer:- it is a chemical change. Because when calcium carbonate is heating it is converting in calcium oxide and carbon dioxide. chemical reaction is occurred here. Thus it is a chemical change.
b) Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.
Answer:-
By dissolving the given products acidicand basic solutions can be prepared as follows:-
Acidic solution:-
CO2+ H2O → H2CO3
Basic Solution:-
CaO + H2O → Ca(OH)2
30) Non-Metals are usually poor conductors of heat and electricity. They are non-lustrous, non- sonorous, non- malleable and are
(a) Name a lustrous non- metal.
Answer:- Iodine Is a lustrous Non- metal.
(b) Name a non- metal which exists as a liquid at room temperature.
Answer:- Bromine is a non-metal and it exists as liquid at room temperature.
(c) The allotropic form of a non- metal is a good conductor of electricity. Name the allotrope.
Answer:- Graphite is good conductor of Electricity but it is a non- metal.
(d) Name a non- metal which is known to form the largest number of compounds.
Answer:- carbon is a non- metal which forms the largest number of compounds.
(e) Name a non- metal other than carbon which shows allotropy.
Answer:- sulphur and phosphorus are non- metals other than carbon which shows allotropy.
(f) Name a non metal which is required for combustion.
Answer:- Oxygen is non- metal which is needed for combustion.
31) Classify the substances given in fig. 2.2 into elements and compounds.
Answer:-
The elements and compounds are classified as follows:-
Elements Compounds
O2 CaCO3
Zn H2O
Hg sand
Cu Wood
Diamond (carbon) NaCl (aq)
F2
33) Which of the following are not compounds?
Chlorine gas
Answer:-it is not Compound
Potassium permanganate
Answer:- it is a Compound
Iron
Answer:- it is Not Compound
Iron sulphide.
Answer:- it is a Compound
Aluminium
Answer:- it is Not compound
Iodine
Answer:- it is Not compound
Carbon
Answer:- It is Not compound
Carbon monoxide
Answer:- it is a Compound
Sulphur powder
Answer:- it is Not compound
Long Answer Questions
33) Fractional distillation is suitable for separation of miscible liquids with a boiling point difference is about 25 k or less. What part of fractional distillation apparatus makes it efficient and possess an advantage over a simple distillation process. Explain using a diagram.
Answer:- Fractional distillation is suitable process for separation of two miscible liquids with a boiling point difference. Fractional column is the important part of the fractional distillation apparatus. It has glass beads in it. These glass beads provides surface to vapours. These high boiling liquids vapours colloids in each other, and thus they lose energy so that they could condensed quickly and distilled. Latent heat released during the boiling of liquid helps to reach the vapours atThe height of the fractionating column.
Process of fractional distillation
The advantages of the fractional distillation are –
A) Both Evaporation and condensation takes place simultaneously during the process.
B) The method of fractional distillation separate the liquids with a boiling point difference of 25k or less than 25k.
C) Petroleum liquids which contains several compounds can be separated by fractional distillation.
34) A) under which category of mixtures will you classify alloys and why?
Answer:-
Alloys will be classified under the homogenous mixture. Because, alloys is an mixture (homogeneous) which is formed by two or more elements. These elements can be two Metals or two Non-Metals. Or a metal with non- metal. It is classified as an homogenous mixture, because it contains two or more properties of elements which form the alloy. It has the uniform constituents.
B) a solution is always a liquid. Comment.
Answer:- No. Solution is not always a liquid. Other solutions are also possible. Solid and gaseous solutions are also possible. For examples – Air and brass.
C) can a solution be heterogeneous?
Answer:- No. Solutions cannot be heterogeneous. An solution is the uniform mixture of two or more substances.
35) Iron fillings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’ , part ‘A’ was heated strongly while part ‘B’ was not heated. Dilute hydrochloric acid was added to both the parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?
Answer:- The hydrogen gas is evolved from both the mixtures.
These is shown by the reaction which is carried out in both the mixtures after adding hydrochloric acid.
In part A
The reaction carried out is
∆
Fe(s) + S (s) –––>FeS
FeS + 2 HCl → FeCl2 + H2S
H2S gas shows that the hydrogen gas has evolved from these reaction.
Now, in part B
The reaction carried out is
Fe(s) + S (s)
Here sulphur does bot react with the ferrous because heat is not provided to it.
after adding diluted HCl in it, it smell so foul like an rotten egg. Thus These shows the formation of H2S gas in it. Also it turns the solution black in colour.
36) A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in fig. 2.3. the filter paper was removed when the water moved near the top of the filter paper.
I) What would you expect to see, if the ink contains three different coloured components ?
Answer:- When the ink contains three different coloured components then we will observe the three different bands.
II) Name the technique used by the child.
Answer:- Chromatography is the technique used by the child.
III) Suggest one more application of this technique.
Answer:- The technique of paper chromatography is used for the separation of the pigment which is present in the chlorophyll.
37) A group of students took an old shoe box and covered it with a black paper From sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/ tumbler in the box as shown in the fig. 2.4 they were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?
a) Explain why the milk sample was illuminated. Name the phenomenon involved.
Answer:- Milk is an colloidal solution. Colloidal solution has a small particles therefore the light scatters and it shows the Tyndall effect. These Tyndall effect results into illuminated milk.
b) Same results were not obtained with a salt solution.
Explain.
Answer:- Salt solution is a homogeneous solution. It contains small particles which does the scatter the light passing through it. Therefore there is not any Tyndall effect.
c) Can you suggest two more solutions which would show the same effect as shown by the milk solution?
Answer:-
Milk is an colloidal solutions which exhibit ls the Tyndall effect, another examples of the colloidal solutions are sulphur Solution and detergent solution.
38) Classify each of the following as a physical or a chemical change. Give reasons.
a) Drying of a shirt in the sun.
Answer:- it is a physical change. Because drying of shirt in the Sun-light does bot contain any chemical reaction in it. It is just due to Evaporation process out in it. Therefore it is an physical process.
b) Rising of hot air over a radiator.
Answer:- It is a physical change. Because it does contain any chemical reaction or chemical change during the rising of hot air over a radiator.
c) Burning of kerosene in a lantern.
Answer:- it is a Chemical change. Burning of kerosene in the lantern shows the chemical change. Burning process needs oxygen and during burning it produces carbon dioxide gas.
d) Change in the colour of black tea on adding lemon juice to it.
Answer:- it is a chemical change. Change in colour shoes the chemical change in the tea. Therefore it is a chemical change.
e) Churning of milk cream to get butter.
Answer:- it is a physical change. Churning of milk cream does not show any chemical reaction or chemical change in it. Here centrifugation process is used for turning the milk into butter.
39) During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water while sarika prepared it by dissolving 10g of sugar in water to make 100g of the solution.
a) Are the two solutions of the same concentration.
Answer:- No the two Solutions are not of same concentration.
The formula for the concentration of soln
Mass % = (Mass of solute/Mass of solute + mass of solvent) × 100
b) Compare the mass % of the two solutions.
Answer:-
The Solution made by the Ramesh is concluded as follows-
Mass % = (10 / 10+100) X 100
= (10/110) × 100
= 9.09%
The Solution made by the sarikahas concluded as follows by using the formula.
Mass % = (10/100) × 100
= 10%
Both have made different mass % of Solution. The Solution made by sarika has higher mass % than ramesh. Ramesh has 9.09% and sarika has 10% mass %.
40) You are provided with a mixture containing sand, iron filings, ammonium chloride and sodium chloride. Describe the procedures you would use to separate these constituents from the mixture?
Answer:-
The provided mixture can be separated by using following procedures step by step-
A) By using the magnet iron fillings can be separated out from the mixture.
B) Further the Sublimation process is carried out for the separation of ammonium chloride from the remaining mixture. It is collected in the inverted china dish.
C) Later on water is added to the mixture the sand it allowed to settle down the beaker it is known as sedimentation.
D) Decantation – the liquid is further decanted in the other beaker.
E) Filtration – Filtration separate the sand over the filter paper.
F) The filtrate water is heated. For evaporate the water. Water gets converted into the vapours and salt remains in it.
41) Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
a) 1.00 g of NaCl + 100g of water
b) 0.11 g of NaCl + 100g of water
c) 0.01g of NaCl+ 99.99g of water
d) 0.10 g of NaCl+ 99.90g of water
Answer:-
Calculation for option A)
1.00
Mass % = 1.00/1.00+100 × 100 = 0.99%
Calculation for option B)
Mass % = 0.11/0.11+100 x 100 = 0.11%
Here, the calculation for C)
= 0.01/0.01+ 99.99 ×100
= 0.01/0.01+ 99.99 ×100
= 0.01/100 × 100
= 0.01 g
Calculation for option D)
Mass% = 0.1/0.1+100 × 100 = 0.1%
Answer is C) = 0.01 g which is correctly shows the mass% Solution of sodium chloride in water. Therefore the C is the correct answer.
42) Calculate the mass of sodium sulphate required to prepare it’s 20% (mass percent) solution in 100g of water ?
Answer:-
Take therequired mass of sodium sulphate= x g
Then the mass of solution will be= (x+100)g
Now let prepare it’s 20% (mass percent)by following formula.
Therefore, the 20% mass of sodium sulphate is 25 g.