Lakhmir Singh Manjit Kaur Chemistry 2019 Solutions for Class 9 Science Chapter 5 Multiple Choice Questions (MCQs) are provided here with simple step-by-step explanations. These solutions for Multiple Choice Questions (MCQs) are extremely popular among Class 9 students for Science Multiple Choice Questions (MCQs) Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Lakhmir Singh Manjit Kaur Chemistry 2019 Book of Class 9 Science Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Lakhmir Singh Manjit Kaur Chemistry 2019 Solutions. All Lakhmir Singh Manjit Kaur Chemistry 2019 Solutions for class Class 9 Science are prepared by experts and are 100% accurate.

Page No 214:

Question 1:

Four students, (A), (B), (C) and (D) observed the colour and solubility of iron, sulphur and iron sulphide in carbon disulphide.  The tick mark (✔) represents 'soluble', and cross mark (✕) represents 'insoluble', in carbon disulphide. Their observations are tabulated below.

Student Colour Solubility in carbon disulphide
Fe S FeS Fe S FeS
(A) Yellow Silvery Greyish silver (✔) (✕) (✔)
Silvery Orange Reddish brown (✕) (✔) (✔)
(B)
(C) Grey Yellow Greyish black (✕) (✔) (✕)
(D) Silvery White Silvery white (✔) (✕) (✕)
The student, who correctly reported the observations, is student :
(1) (A)
(2) (B)
(3) (C)
​(4) (D)


 

Answer:

Iron is of grey colour and sulphur is of yellow colour. When iron and sulphur react they form black coloured iron sulphide.
Sulphur is only soluble in CS2 solution. So, out of four students, only student C correctly reported the observation. 

Hence, the correct answer is option 3.

Page No 214:

Question 2:

The white of an egg, common salt, sugar and fine sand are added to water separately in beakers as shown below. The mixture is stirred well. A suspension will be formed in the beaker :

(1) I
(2) II
(3) III
(4) (IV)

Answer:

A heterogeneous mixture in which solute particles do not dissolve, but left suspended in the solution is known as suspension. So, out of given mixtures, I, II and III will completely mix in the solvent while only IV having sand will remain suspended in the solution.

Hence, the correct answer is option 4.
 

Page No 214:

Question 3:

The correct procedure of heating iron-sulphur mixture to prepare iron sulphide is :
(1) heat the powder mixture at the base of the test tube using a blue flame throughout.
(2) heat the iron filings and sulphur mixture in the middle of the test tube using yellow flame throughout.
(3) heat the powder mixture at the top of the test tube using an orange flame throughout.
(4) heat the iron fillings-sulphur mixture at 3/4 quarters of the test tube using a red flame throughout.

Answer:

Fe and S mixture is prepared by heating them in the test tube under blue flames, they are heated until the grey-black coloured iron sulphide is obtained.

Hence, the correct answer is option 1.

Page No 214:

Question 4:

A student while heating solid lead nitrate taken in a test tube would observe :
(1) white residue of PbO2
(2) green residue of NO2
(3) yellow residue of PbO
(4) brown residue of NO

Answer:

When lead nitrate is heated, a yellow residue of lead oxide is obtained and brown colored gas is liberated i.e. nitrogen dioxide.
2Pb(NO3)22PbO+4NO2+O2

Hence, the correct answer is option 3.

Page No 214:

Question 5:

The following precautions were listed for the experiment on the determination of the melting point of ice. The incorrect precaution is :
(1) The bulb of the thermometer should be kept surrounded with crushed ice.
(2) Ice should be stirred regularly to keep a uniform temperature throughout.
(3) The final temperature should be noted by keeping the eyes in line with the level of mercury.
(4) Only the tip of the bulb of the thermometer should just touch the crushed ice.

Answer:

The temperature of any substance/material is determined by the bulb of the thermometer. So, for correct measurement of temperature, the bulb of thermometer should be properly surrounded with crushed ice.

Hence, the correct answer is option 4.

Page No 214:

Question 6:

The correct procedure for preparing a colloidal solution of egg albumin in water is :
(1) to break the egg shell, take only the white portion and to add it to water with constant stirring.
(2) to break the egg shell, take only the yellow portion and to add it to boiling water with constant stirring
(3) to boil the egg first, to break the egg shell, to add the white portion to ice cold water and to mix.
(4) to boil the egg first, to break the egg shell, to add the yellow portion to water and to mix.

Answer:

The correct procedure to prepare a colloidal solution of egg albumin and water is to break the eggshell first, after that only while part of the egg should be removed and carefully added to the water. Constant stirring is also required for proper mixing of egg and water.

Hence, the correct answer is option 1.



Page No 215:

Question 7:

Four students (A), (B), (C) and (D) independently observed the evaporation of water under different conditions, and recorded the temperature of water at regular intervals as shown below.

Student Placing of experimental set up in/under Temperature recording for 15 minutes
(A) sun increased gradually
(B) open air decreased gradually
(C) a fan initially increased, then became constant
(D) a corner of the room  initially increased, then gradually decreased
The correct recording of observations is that of the student :
(1) (A)
(2) (B)
(3) (C)
​(4) (D)

Answer:

In the open air, temperature and pressure of the environment will remain the same, so first evaporation increases but gradually it will decrease. So, after 15 minutes temperature observation will be gradually decreasing.

Hence, the correct answer is option 2.

Page No 215:

Question 8:

A student takes a mixture of sand and ammonium chloride in a china dish and heats it under a funnel fitted with a plug over a flame. He would observe that :
(1) solid sand gets deposited on the lower cooler parts of the funnel while solid ammonium chloride remains in the china dish.
(2) sand and ammonium chloride get deposited on hotter parts of the funnel.
(3) ammonium chloride gets deposited on the cooler parts of the funnel and sand remains in the china dish.
(4) sand collects on cooler parts of the funnel while ammonium chloride melts in the china dish.

Answer:

Solid ammonium chloride directly converts to gas phase on heating, and these vapours are then collected at the cooler parts of the funnel, while sand remains on the china dish. This conversion of solid directly into the gaseous phase in known as sublimation.

Hence, the correct answer is option 3.

Page No 215:

Question 9:

A student takes some water in a beaker and heats it over a flame for determining its boiling point. He keeps on taking its temperature readings. He would observe that the temperature of water :
(1) keeps on increasing regularly
(2) keeps on increasing irregularly
(3) first increases slowly, then decreases rapidly and eventually becomes constant
(4) first increases gradually and then becomes constant.

Answer:

When we boil the water, first its temperature will rise gradually but after some time it becomes constant. During phase transformation temperature of the water becomes constant because all the heat supplied to increase the temperature will be utilizing in changing the state by overcoming the force of attraction between the particles.

Hence, the correct answer is option 4.

Page No 215:

Question 10:

Which of the following is the correct set of apparatus to separate common salt and sand by filtration process :
 
(1) I
(2) II
(3) III
(4) IV

Answer:

For filtration of common salt and sand first setup is preferred. When water is poured with the help of glass rod so that pressure of water decreases and it will easily pass through the filter paper and sand will remain on the funnel and salt will pass through it.

Hence, the correct answer is option 1.

Page No 215:

Question 11:

A student added the following substances to water kept in four separate beakers. He stirred the mixture well and filtered each one of them through a filter paper. He would obtain a solid residue on the filter paper in the case of :
(1) egg albumin
(2) common salt
(3) chalk powder
(4) alum

Answer:

Chalk powder (chemically known as calcium carbonate) is insoluble in water as it forms a precipitate in water and does not dissolve, so chalk powder will be obtained from filter paper.

Hence, the correct answer is option 3.
 

Page No 215:

Question 12:

In an experiment, carbon disulphide was added to a test-tube containing a mixture of iron filings and sulphur powder as shown in the given diagrams :

The correct observation is represented in diagram :
(1) I
(2) II
(3) III
(4) IV

Answer:

Addition of carbon disulphide to a mixture containing iron filings and sulphur powder leads to the formation of a clear yellow solution when sulphur powder dissolves in carbon disulphide, on gentle shaking. Iron fillings being insoluble settle in the bottom. These can be separated by filtration. When the solution is allowed to evaporate, the powder of solid sulphur is obtained.

Hence, the correct answer is option 3.
 



Page No 216:

Question 13:

The colour of insoluble product formed when sodium sulphate solution and barium chloride solution are mixed together is :
(1) blue
(2) yellow
(3) white
(4) red-brown 

Answer:

When barium chloride reacted with sodium sulphate solution then double displacement reaction occurs. A white precipitate of barium sulphate and the aqueous solution of sodium chloride is obtained. The reaction involves is as follows:
BaCl2(aq)+Na2SO4(aq)Ba(SO4)2(s)White precipitate+NaCl(aq)

Hence, the correct answer is option 3.

Page No 216:

Question 14:

A student, by mistake, mixed sulphur powder with iron filings. The following techniques were suggested to separate the sulphur from the mixture out of which he has to choose one :
A. dissolving in carbon disulphide, filtration, evaporation
B. dissolving in water at room temperature and filtration
C. dissolving in hot water, filtration and evaporation
D. dissolving in ice cold water and filtration
The correct technique is :
(1) A
(2) B
(3) C
(4) D

Answer:

Addition of carbon disulphide to a mixture containing iron filings and sulphur powder leads to the formation of a clear yellow solution when sulphur powder dissolves in carbon disulphide, on gentle shaking. Iron fillings being insoluble settle in the bottom. These can be separated by filtration. When the solution is allowed to evaporate, the powder of solid sulphur is obtained.

Hence, the correct answer is option 1.

Page No 216:

Question 15:

A student has done the labelling for the experimental set-up for separating a mixture of sodium chloride and camphor as indicated in the diagram given here. The parts/substances that have been incorrectly labelled are :

(1) I, III, VIII 
(2) II, III, VII
(3) I, II, VIII
(4) III, V, VII

Answer:

In the given figure, label "I" is represented as cork which is wrong, it is a cotton plug. 
Label "II" is represented as separating funnel, it is an inverted funnel, not a separating funnel.
Label "VIII" is represented as pure NaCl, but it is sublimate of camphor.

Hence, the correct answer is option 3. 
 

Page No 216:

Question 16:

A teacher gave an impure sample of alum containing fine sand as impurity to a student. He asked him to recover pure alum from this sample. The correct procedure to be followed would be to :
(1) dissolve the impure sample of alum in water, filter and evaporate the filtrate
(2) dissolve the impure sample of alum in alcohol and filter
(3) move a magnet over the impure sample of alum
(4) dissolve the impure sample of alum in carbon disulphide, filter and evaporate the filtrate

Answer:

To prepare the pure alum, first, dissolve the impure alum in water than filter it. All the insoluble impurities remain in the filter paper. Then evaporate the water and pure alum will be obtained.

Hence, the correct answer is option 1.

Page No 216:

Question 18:

A student took some lead nitrate compound in a boiling tube and heated it strongly. The gas/gases evolved on heating this compound is/are :
(1) NO2
(2) NO+ CO2
(3) NO2 + O2
(4) N+ O2

Answer:

When lead nitrate is heated it splits into simpler substances, 
2Pb(NO3)2Colourless2PbOYellow ppt+4NO2Brown fumes+O2
So, NO2 and O2 are liberated from the reaction.

Hence, the correct answer is option 3.

Page No 216:

Question 19:

When a student heated a colourless solid compound, then brown fumes of a gas are evolved. The colourless solid is most likely to be :
(1) ferrous sulphate
(2) lead carbonate
(3) lead nitrate
(4) lead sulphate

Answer:

When lead nitrate is heated it splits into simpler substances, 
2Pb(NO3)2Colourless2PbOYellow+4NO2Brown+O2
NOgas imparts brown fumes to the solution.

Hence, the correct answer is option 3.



Page No 217:

Question 17:

The colour of residue left behind on heating lead nitrate when it is still hot is :
(1) green
(2) yellow
(3) black
(4) reddish-brown

Answer:

When lead nitrate is heated it decomposes into simpler substances, they impart different colour to the solution.
2Pb(NO3)2colourless2PbOyellow ppt+4NO2brown fumes+O2
Due to the formation of PbO, it shows yellow colour.

Hence, the correct answer is option 2.

Page No 217:

Question 20:

When dilute sulphuric acid is added to granulated zinc placed in a test-tube, the observation made is :
(1) the surface of the metal turns shiny
(2) the reaction mixture turns milky
(3) odour of sulphur dioxide is observed
(4) a colourless and odourless gas evolves with bubbles

Answer:

When zinc reacted with dilute hydrochloric acid, it gives zinc chloride and hydrogen gas is evolved, hydrogen gas is colourless and odourless gas which evolves with bubbles.
Zn(s)+2HCl(aq)ZnCl2(aq)+H2(g)

Hence, the correct answer is option 4.

Page No 217:

Question 21:

What is the correct order of the methods you would apply to separate the components of a mixture of ammonium chloride, common salt and sand ?
(1) dissolving in water, filtration, evaporation and sublimation
(2) dissolving in water, evaporation and sublimation
(3) sublimation, dissolving in water, filtration and evaporation
(4) moving a magnet, dissolving in water and sublimation

Answer:

Solid ammonium chloride was directly converted to the gas phase, and can easily be separated out from the mixture. Common salt dissolved in water can also be separated from the mixture by the evaporation method.
Sand easily be separated by the filtration method.

Hence, the correct answer is option 3. 

Page No 217:

Question 22:

Four students took separately the mixture of sand, common salt and ammonium chloride in beakers, added water, stirred the mixture well and then filtered. They reported their observations as shown below.

Student As residue In the filtrate
I Ammonium chloride Sand Common salt
II Common salt, Sand Ammonium chloride
III Sand, Ammonium chloride Common Salt
IV Sand Ammonium chloride, Common salt
Who reported the observations in the correct order of the components as residue and in the filtrate ?
(1) I
(2) IV
(3) III
(4) II

Answer:

Solid ammonium chloride directly converted to the gas phase, and can easily be separated out from the mixture as residue
Common salt dissolves in water and after evaporation, it can also be separated from the mixture as residue.
So ammonium chloride and common salt are categorized as residue while sand was left in the mixture as filtrate.
So, the observations made by student III were correct.

Hence, the correct answer is option 3.

Page No 217:

Question 23:

When zinc metal reacts with dilute sulphuric acid, a gas is evolved. Which one is a correct statement about the nature of this gas ?
(1) colourless with suffocating odour
(2) reddish brown and odourless
(3) colourless and sweet smelling
(4) colourless and odourless

Answer:

When zinc reacts with dilute hydrochloric acid, it gives zinc chloride and hydrogen gas evolves. Hydrogen is colourless and odourless gas which evolves with bubbles.
Zn(s)+2HCl(aq)ZnCl2(aq)+H2(g)

Hence, the correct answer is option 4.

Page No 217:

Question 24:

The reaction between iron and copper sulphate solution represents which type of reaction ?
(1) decomposition
(2) combination
(3) single displacement
(4) double decomposition

Answer:

As iron is more reactive than copper, so it will displace copper from copper sulphate solution and form iron sulphate.
Fe(s)+CuSO4(aq)FeSO4(aq)+Cu(s)
So, the following reaction is an example of a displacement reaction.

Hence, the correct answer is option 3.

Page No 217:

Question 25:

When a burning magnesium ribbon is introduced in a gas jar containing oxygen, it will burn with :
(1) a reddish flame
(2) a pale blue flame
(3) a golden yellow flame
(4) a dazzling white flame

Answer:

Magnesium has a high affinity for oxygen, therefore magnesium reacts with oxygen to form magnesium oxide.
2Mg+O22MgO
This formation of MgO imparts dazzling white flame.

Hence, the correct answer is option 4.
 

Page No 217:

Question 26:

A student heats calculated amounts of iron filings and sulphur powder together in a boiling tube. He will obtain :
(1) a homogeneous mixture of Fe and S
(2) a heterogeneous compound of Fe and S
(3) a homogeneous compound of FeS
(4) a heterogeneous mixture of FeS, Fe and S

Answer:

When iron and sulphur are heated together in a boiling tube, they react together to form a grey-black compound of iron sulphide.  The mixture formed will be homogeneous i.e. two elements mix together thoroughly and cannot be separated.
Fe+SFeS

Hence, the correct answer is option 3.

Page No 217:

Question 27:

Which of the following mixture can be separated completely by the process of sublimation ?
(1) fine sand and cane sugar
(2) sodium chloride and potassium permanganate
(3) potassium chloride and ammonium chloride
(4) barium chloride and sodium sulphate

Answer:

When the solid phase directly converts to gas phase on heating, it is known as sublimation. So, among the following options, only potassium chloride and ammonium chloride can be separated by sublimation as ammonium chloride directly converts to the gas phase on heating.

Hence, the correct answer is option 3.

Page No 217:

Question 28:

Four set ups as given here were arranged to identify the gas evolved when dilute sulphuric acid was added to zinc granules.The most appropriate set up is :
 
(1) I
(2) II
(3) III
(4) IV

Answer:

When zinc reacts with dilute sulphuric acid then it gives zinc sulphate and hydrogen gas will evolve.
Zn(s)+H2SO4(aq)ZnSO4(aq)+H2(g)
So, from the following given setups in setup 4 hydrogen gas will evolve easily as there is proper space for the gas to evolve.

Hence the correct option is 4.



Page No 218:

Question 29:

An iron nail was dropped into copper sulphate solution. After some time, the colour of the solution changed from :
(1) light green to blue
(2) blue to light green
(3) light green to colourless
(4) blue to yellow

Answer:

Blue coloured copper sulphate solution on reaction with iron gives light green coloured iron sulphate.
CuSO4(aq)Blue+Fe(s)GreyFeSO4(aq)Green+Cu(s)Brown

Hence, the correct answer is option 2.

Page No 218:

Question 30:

A student was asked by her science teacher to prepare a true solution. She dissolved the solute in water but forgot to record its name. What may be the correct name of the solute ?
(1) barium sulphate
(2) sulphur powder
(3) alum
(4) egg albumin

Answer:

The true solution is a homogeneous mixture of two or more compounds in which the size of the solute particle is less than 1nm.
Barium sulphate is insoluble in water, while egg albumin and sulphur powder form colloidal sols, but alum dissolves in water to give the true solution.

Hence, the correct answer is option 3.

Page No 218:

Question 31:

A student placed a clean iron nail in blue coloured copper sulphate solution for a considerable time. He observes that :
(1) iron nail gets green coating
(2) iron nail gets brown coating
(3) iron nail gets no coating
(4) iron nail gets blue coating

Answer:

When a clean iron nail is placed in a blue coloured copper sulphate solution, it displaces the copper from copper sulphate being more reactive than copper. The displaced brown coloured copper gets deposited over the clean iron nail, and with time, it forms a brown coloured coating on the iron nail. 
CuSO4(aq)Blue+Fe(s)GreyFeSO4(aq)Green+Cu(s)Brown

Hence, the correct answer is option 2.

Page No 218:

Question 32:

When a student added a few drops of barium chloride solution to sodium sulphate solution, he obtained a white precipitate instantly. Which of the following type of chemical reaction has been carried out by the student ?
(1) combination
(2) double displacement
(3) displacement
(4) decomposition
 

Answer:

When barium chloride is reacted with sodium sulphate then a white precipitate of barium sulphate and aqueous solution of sodium chloride salt is obtained.
BaCl2(aq)+Na2SO4(aq)Ba(SO4)2(aq)+NaCl(aq)
So, the following reaction is an example of double displacement reaction.

Hence, the correct answer is option 2.

Page No 218:

Question 33:

A student was asked to carry out a chemical reaction by placing four different metal strips in CuSO4 solution for a considerable time, one by one. Which of the following metal strip will turn the blue CuSO4 solution to a light green solution in due course of time ?
(1) Fe
(2) Au
(3) Mg
(4) Ag

Answer:

Copper sulphate solution is a blue colour solution, which on reaction with iron gives a green coloured solution of iron sulphate.

CuSO4(aq)Blue+Fe(s)GreyFeSO4(aq)Green+Cu(s)Brown

Hence, the correct answer is option 1.

Page No 218:

Question 34:

We can show that iron is more reactive than copper :
(1) by preparing copper sulphate solution and dipping iron strip in it 
(2) by dipping both the strips in water for some time
(3) by preparing iron sulphate solution and dipping copper strip in it
(4) by heating both iron and copper strips

Answer:

Iron is more reactive metal then copper, so iron displaces copper from blue coloured copper sulphate solution and forms light green coloured iron sulphate solution. 
CuSO4(aq)Blue+Fe(s)GreyFeSO4(aq)Green+Cu(s)Brown
So, we prepare copper sulphate solution and dip an iron nail in it, it will give us easily observable results.

Hence, the correct answer is option 1.

Page No 218:

Question 35:

A student sets up an apparatus to determine the melting point of ice. He takes a beaker half filled with crushed ice and dips a mercury thermometer with an initial reading of room temperature (25ºC) in such a way that the bulb of the thermometer is surrounded by ice. The correct observation obtained by the student is that :
(1) Mercury in the thermometer keeps on falling till it reads, −1ºC and becomes constant thereafter
(2) temperature falls, reaches 0ºC, then it remains constant even after the whole of the ice has melted
(3) the temperature falls in the beginning but starts rising as soon as the ice starts melting
(4) temperature falls, reaches 0ºC and remains constant only as long as both ice and water are present in it
 

Answer:

When the student places the thermometer in a beaker filled with crushed ice, its temperature will decrease until it reaches 0º and then it will become constant. At 0ºC, all of the heat will get utilized in the phase transition of ice to water. So, the temperature of the thermometer will remain constant until all the ice melts.

Hence, the correct answer is option 4.

 



View NCERT Solutions for all chapters of Class 9