Category: Class 11 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “States of Matter – Gaseous State”.

1. What is the lowermost layer of the earth?
a) stratosphere
b) troposphere
c) ionosphere
d) mesosphere
Answer: b
Clarification: It is the lowest layer of the earth’s atmosphere and is held to the earth by the gravitational force. It is very vital for human life and it protects us from harmful radiation. It contains important molecules like dioxygen, carbon dioxide, water vapor, etc.

2. Which of the following statement is true regarding gases?
a) gases are highly incompressible
b) gases exert equal pressure on each and every direction
c) its volume and shape is fixed
d) gases have the highest density among the 3 States of matter
Answer: b
Clarification: The correct statement is that gases exert equal pressure on each and every direction. Among the given options the corrected statements of other options are that gases are highly compressible, they occupy the shape and volume of the container as they have no fixed shape and volume and also that they have the least density among the three states of matter.

3. Which of the following element is not a gas?
a) Hydrogen
b) Oxygen
c) Mercury
d) Nitrogen
Answer: c
Clarification: Mercury is not a gas, it is a liquid in room temperature and it is a metal. There are 11 gases which are gases at room temperature they are hydrogen, Nitrogen, Oxygen, fluorine, chlorine, Helium, Neon, Argon, Krypton, Xenon, and Radon.

4. Gases have low density than that of solids and liquids because of __________
a) no thermal energy
b) higher intermolecular energy
c) both intermolecular energy and thermal energy are the same
d) higher thermal energy
Answer: d
Clarification: In gases, there is less amount of intermolecular energy and higher amount of thermal energy. As we know that thermal energy separates some molecules from one another so gases have low density than that of solids and liquids.

5. Gases mix properly without any mechanical aid.
a) true
b) false
Answer: a
Clarification: As the forces of interaction between molecules of a gas is negligible when compared with solids and gases. They mix properly because of higher thermal energy and lower intermolecular energy, so the above statement is true.

6. Which of the following is not a gas law?
a) Boyle’s law
b) Charles law
c) Hooks law
d) Gay lussac’s law
Answer: c
Clarification: Boyle’s law is about the relationship between pressure and volume while Charles law about temperature and volume. Gay lussac’s law is about pressure-temperature relationship & hooks law is a law that is in Physics relating to stress.

7. What is the percentage of Nitrogen in the atmosphere approximately?
a) 78.09
b) 21
c) 20
d) 32
Answer: a
Clarification: That composition of earth’s atmospheric gases is as follows; 78.09 percent of Nitrogen, 24.95 percent of oxygen, 0.93 percent of argon, 0.04 percentage of carbon dioxide and a small amount of water vapor and other gases in the atmosphere.

8. What can you say about particles motion in gases?
a) only vibratory
b) very slow
c) both vibratory and irregular
d) too Rapid and random
Answer: d
Clarification: A particle’s motion in the gaseous state is too rapid and random while in solids it’s restricted to vibratory motion and in liquids, it’s very slow. This is one of the very basic properties of substances in the gaseous state.

9. How many moles of oxygen are present in 64 grams of oxygen?
a) three moles
b) two moles
c) one mole
d) 16 moles
Answer: b
Clarification: As we know that the number of moles of a gas is given by the amount of the substance in weight divided by the molecular weight of the substance. So in case of oxygen, it is 64 grams divided by 32 grams and that is two moles.

10. At STP conditions how much volume does one mole of a gas comprise of _________
a) 22.4 liters
b) 24 liters
c) depends on the molecular weight of the gas
d) depends on some other conditions
Answer: a
Clarification: Every one Mole of gas at STP consists of 22.4 liters of volume, that is at 0° Celsius of temperature and one-atmosphere pressure or 76 mm of Mercury pressure. Also, note that one mole of a gas is the amount of gas in weight divided by the molecular weight of the gas.

Chemistry Multiple Choice Questions on “Thermodynamics Applications”.

1. When an ideal gas is compressed in a piston using 5 atm of pressure through a 50-metre cube of volume, what is the amount of work done?
a) 10 Newton metre
b) 0.1 Newton metre
c) 250 Newton metre
d) 55 Newton metre
Answer: c
Clarification: When a pressure P is exerted through a volume V the work done is given by P ΔV, so here pressure is 5 atm and volume is a 50-metre cube. The work that is done is 5 atm X 50-metre cube = 250 Newton metre.

2. When the pressure of 3 atm is exerted over a surface area of a 10-metre square, what is a force that is applied?
a) 30 Newton
b) 3.33 Newton
c) 0.33 Newton
d) 0.3 Newton
Answer: a
Clarification: Force is defined as the product of pressure and the surface area so here as pressure is 3 atm and surface area is 10-metre square. The force that is applied equals 3 atm x 10-metre square = 30 Newton.

3. Expansion of gas under zero pressure is free expansion.
a) True
b) False
Answer: a
Clarification: The expansion of a gas in a vacuum without pressure is called free expansion. During the free expansion of gas, the work is not done whether the process is reversible or Irreversible. So the above statement is considered to be true.

4. 6 litres of an ideal gas expands isothermally at a temperature of 300 Kelvin up to 10 litres at a pressure of 5 atm, what is the work done?
a) 30 Newton metre
b) 80 Newton metre
c) 50 Newton metres
d) 20 Newton metre
Answer: d
Clarification: The expression for work done is given by pressure x volume difference, here an ideal gas has a volume difference of 4 litres at 5 ATM pressure. So the work done = 10 – 6 liters x 5atm = 20 Newton metre.

5. Which of the following is an intensive property?
a) Volume
b) Colour
c) Enthalpy
d) Internal energy
Answer: b
Clarification: An intensive property does not depend on the quantity or size of the object, whereas extensive property depends on the quantity and size of the object. Here volume, enthalpy and internal energy are extensive properties, while colour is an intensive property.

6. The value of the product of a universal gas constant and the temperature difference is given by 10 kJ/mol at 1 mole and the internal energy is given by 20 KJ, what is the enthalpy of this system in KJ?
a) 30
b) 10
c) 20
d) 200
Answer: a
Clarification: We know that ΔH = ΔU + nRΔT; where ΔH is the enthalpy, ΔU is the internal energy, n is the number of moles, R is the universal gas constant and ΔT is the temperature difference. So enthalpy is 10 KJ + 20KJ = 30 KJ.

7. What is the difference in heat capacities at constant volume and pressure?
a) Universal volume constant
b) Universal gas constant
c) Universal pressure constant
d) Universal temperature constant
Answer: b
Clarification: We all know that ΔH = ΔU + nRΔT; ΔH = nC_PΔT and ΔU = nC_vΔT; ΔH – ΔU = nC_PΔT – nC_vΔT = n(C_P – C_V) ΔT. By equating L.H.S. and R.H.S., we get n(C_P – C_V) ΔT = nRΔT; C_P – C_V = R. Hence it’s proven that difference of heat capacities at constant volume and pressure is the universal gas constant.

8. Write temperature difference in terms of heat capacity and heat energy?
a) ΔT = q/C
b) ΔT = qC
c) ΔT = C/q
d) ΔT = qm/C
Answer: a
Clarification: As we know that q = CΔT, where q is the heat energy, C is the specific heat and ΔT is the temperature difference, When the temperature difference is expressed in terms of the heat capacity and heat energy, it is given as ΔT = q/C.

9. If gas is expanded freely from 1 litre to 5 litres at a temperature of 60-degree centigrade what is the work done?
a) Positive
b) Negative
c) 0
d) Infinity
Answer: c
Clarification: When gas is expanded freely in a vacuum there is zero pressure exerted. The formula for work done is given by pressure X change in volume = 0 x 4. So the work done is zero in the process of free expansion.

10. The specific heat at constant pressure is given by the expression ____________
a) C_V = dq/dT
b) C_P = dq/dT
c) C_V = dqdt
d) C_P = dq/dt
Answer: b
Clarification: As we know that dq = CdT; where q is the heat energy, C is the specific heat and T is the temperature. At constant pressure, specific heat is given as C_P. The specific heat at constant pressure is given by the expression C_P = dq/dt.

Chemistry Question Papers for Class 11 on “Equilibrium – Buffer Solutions”.

1. Carbonic acid and sodium bicarbonate are present in blood as a buffer.
a) true
b) false
Answer: a
Clarification: A solution that resists the change in its PH value by the addition of a small amount of acid or base is called a buffer solution. The buffer system present in the blood is carbonic acid and sodium bicarbonate.

2. Which of the following is not an acidic buffer?
a) Acetic Acid-Sodium acetate
b) Boric acid-borax
c) Ammonium hydroxide-ammonium chloride
d) All are acidic buffers
Answer: c
Clarification: An acidic buffer has a pH value of less than 7, Acetic Acid-Sodium Acetate and boric acid-borax are examples of acidic buffers, but ammonium Hydroxide-ammonium chloride has a pH of greater than 7, so they are basic buffers.

3. Which of the following is an equation used to calculate the pH of a buffer solution for an acidic buffer?
a) pH = pK_a + log[salt]/[acid]
b) pOH = pK_a + log[salt]/[acid]
c) pH = pK_b + log[salt]/[acid]
d) pH = pK_a + log[salt][acid]
Answer: a
Clarification: Equation that is used to calculate the pH of a buffer solution for an acidic buffer is pH = pK_a + log[salt]/[acid]. This equation is known as henderson-hasselbalch equation, it is used for making of buffer solutions.

4. Buffer solution is destroyed when _____________
a) addition of weak base
b) addition of strong acid or base
c) addition of weak acid
d) addition of a salt
Answer: a
Clarification: If the addition of a strong acid or base changes the pH of a buffer by unit, the buffer solution is assumed to be destroyed that is new pH = pK_a ± 1; that means [salt]/[acid] or [acid]/[salt] = 10 or 1/10.

5. What is the buffer capacity if 3 moles are added in 5 litres of the solution to change the pH by 2 units?
a) 0.2
b) 0.5
c) 0.15
d) 0.3
Answer: d
Clarification: Buffer capacity is defined as the number of moles of acid or base added in one litre of the solution to change the pH by Unity. Therefore here buffer capacity = 3/5 divided by 2 = 0.6/2 = 0.3. The buffer capacity is given as 0.3.

6. Buffer capacity of a buffer is given as two units for a change in pH by Unity. Then what is the number of moles of acid or base, added in one litre of the solution?
a) 2
b) 0.5
c) 1
d) 4
Answer: a
Clarification: Buffer capacity is denoted by Φ = number of moles of acid or base added to one litre of the buffer by a change in pH. Here the change in pH is given by 1 and the buffer capacity is given by 2, therefore by substituting, we get that 2 moles of acid or base are added in one litre of the solution.

7. If 0.20 mol/L CH₃COOH and 0.50 mol/L CH₃COO^– together make a buffer solution, calculate the pH of the solution if the acid dissociation constant of CH₃COOH is 1.8 × 10^-5.
a) 2.09
b) 5.14
c) 2.65
d) 3.98
Answer: b
Clarification: We have henderson-hasselbalch equation as pH = pK_a + log[salt]/[acid]. So by substituting the concentrations of silent and acid along with the acid dissociation constant, we get pH = -log[1.8 × 10^-5] + log [0.50mol/L]/[0.20mol/L] = 5.14.

8. Note that the pK_a here is given by 4.752, a buffer is made using 0.8 M acetic acid and 1 M Sodium Acetate what do you think its pH is(log10/8 = 0.097)?
a) 4.84
b) 4.85
c) 4.849
d) 4.846
Answer: b
Clarification: According to the Henderson hasselbalch equation pH = pK_a + log[salt]/[acid], if we substitute the concentration of salt as 1 M and the concentration of acid as 0.8 M, pH = 4.752 + 0.097 = 4.849 is the required answer.

9. If the pH of a substance is given by 3 then what is the pOH of the substance?
a) 3
b) 7
c) 14
d) 11
Answer: d
Clarification: We know that the sum of the pH and pOH of any substance is equal to 14 that is pH + pOH = 14. So here the pH of a substance is given by 3 the pH of the substance = 14 – 3 = 11, 11 is the required answer.

10. Which of the following do you think is a correct statement?
a) Ammonium hydroxide / ammonium chloride is an acidic buffer
b) boric acid / borax is an acidic buffer
c) henderson hasselbalch equation is given by pH = pK_b + log[salt]/[acid]
d) PH + pOH = 4
Answer: b
Clarification: Ammonium hydroxide / ammonium chloride is a basic buffer, henderson hasselbalch equation is given by pH = pK_a + log[salt]/[acid] and pH + pOH = 14. So the only correct statement is that boric acid / borax is an acidic buffer.

To practice Chemistry Question Papers for Class 11,

Chemistry Multiple Choice Questions on “s-Block Elements – Group 2 Elements : Alkaline Earth Metals”.

1. Which of the following is not an alkaline earth metal?
a) beryllium
b) boron
c) aluminium
d) calcium
Answer: a
Clarification: Beryllium, belongs to the 2^nd group. It is not called as an alkaline earth metal because the first element of this group that is beryllium is different from the rest of the members and it also shows a diagonal relationship with aluminium.

2. How many electrons do group 2 elements have in their S orbital of the valence shell?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: The alkaline earth metals have two electrons in the s-orbital of their valence shell and their general electronic configuration as represented as ns². Similar to the alkali metals these elements compounds are predominantly ionic.

3. Which of the following order is correct with respect to the hydration enthalpy?
a) B > Mg > Ca < Sr > Ba
b) Be⁺² > Mg⁺² > Ca⁺² > Sr⁺² > Ba⁺²
c) B > Mg < Ca > Sr > Ba
d) B > Mg > Ca > Sr < Ba
Answer: b
Clarification: The hydration enthalpy decreases with the increase in ionic size along with the group towards down, the correct order of hydration enthalpy is given as Be⁺² > Mg⁺² > Ca⁺² > Sr⁺² > Ba⁺² and the hydration enthalpies of alkaline earth metal ions are greater than the size of alkali metal ions. We can say that these are extensively hydrated than them.

4. What is the colour of barium?
a) brick red
b) crimson
c) apple green
d) blue
Answer: c
Clarification: The colours of Barium, Strontium and Calcium are Apple Green, Crimson and brick red. These are the colours of their flames, the colours occur when an electron is excited and jumps into a higher energy level and then drop back. They emit the radiation in the form of visible light.

5. Powdered beryllium burns in order to give ___________
a) barium sulphate
b) beryllium chloride
c) beryllium nitride
d) beryllium hydride
Answer: c
Clarification: We know that beryllium is kinetically inert to Oxygen and water as it forms oxide film on the surface. But whereas powdered beryllium burns brightly on ignition in the air in order to give the oxides and nitrides of beryllium.

6. Which of the following is the best route to prepare BeF₂?
a) thermal decomposition of BeF₂
b) thermal decomposition of beryllium
c) thermal decomposition of (NH₄)₂BeF₄
d) thermal decomposition of Barium sulphate
Answer: c
Clarification: All the alkali Earth metals combine with halogen at elevated temperatures forming their halides, the thermal decomposition of (NH₄)₂BeF₄, is best road and to prepare BeF₂ and BeCl₂ is conveniently made from the oxide.

7. Can beryllium hydride be prepared by combining with hydrogen through Heating?
a) Yes
b) No
c) Maybe
d) May not be
Answer: b
Clarification: All the alkaline earth metals (except beryllium) their hydrides can be prepared by combining with hydrogen. We can prepare beryllium hydride, through heating beryllium chloride with lithium aluminium hydride.

8. The reduction potential of alkaline earth metals is ___________ alkali metals.
a) may be equal to
b) greater than
c) less than
d) equal to
Answer: c
Clarification: Although the alkaline earth metals are reductants in nature, their reducing potential is not as greater as alkali metals. Beryllium has a negative value when compared to the other elements in its group, this is due to the large hydration energy associated with the small size of beryllium.

9. Which of the following is a component of milk of magnesia?
a) magnesium oxide
b) magnesium sulphate
c) magnesium hydroxide
d) magnesium chloride
Answer: c
Clarification: Suspension of magnesium Hydroxide in water is known as milk of magnesia and it is used as an antacid in order to treat acidity. It works as a strong base while magnesium carbonate is an ingredient and toothpaste.

10. With which of the following elements magnesium does not form an alloy?
a) manganese
b) aluminium
c) zinc
d) barium
Answer: d
Clarification: Magnesium combines with few elements like aluminium, manganese, zinc and tin in order to form a lot of magnesium aluminium alloys which are very light in mass and are used in the construction of aircraft. There are also many other uses of magnesium.

Chemistry Multiple Choice Questions on “Qualitative Analysis of Organic Compounds”.

1. Carbon and hydrogen are detected by heating the compound with which of the following?
a) Copper (II)oxide
b) Iron(II)oxide
c) Iron(III)oxide
d) Copper(I)oxide
Answer: a
Clarification: Carbon and hydrogen are detected by heating the compound with copper (II) oxide. In this way, the carbon that is present in the compound will be oxidized to carbon dioxide and hydrogen to water. Carbon dioxide is tested by limewater which develops turbidity, and hydrogen is tested with anhydrous copper sulphate, which turns blue.

2. Which compound gets precipitated in the detection of carbon and hydrogen?
a) Copper
b) Carbon dioxide
c) Calcium carbonate
d) Copper sulphate
Answer: c
Clarification: Calcium carbonate (CaCO₃) gets precipitated during the detection of carbon and hydrogen. Firstly carbon will react with copper (II) oxide to form copper and carbon dioxide. Then, hydrogen will react with the same copper (II) oxide to form copper and water. After this, the presence of carbon dioxide is tested by reacting it with calcium hydroxide and the product obtained is calcium carbonate, which gets precipitated and water.

3. Identify the element that cannot be detected by Lassaigne’s test.
a) Nitrogen
b) Fluorine
c) Sulfur
d) Phosphorous
Answer: b
Clarification: Fluorine cannot be detected by lassaigne’s test. Even though lassaigne’s test is used for the detection of halogens, all halogens except fluorine can be detected. This is because, in lassaigne’s test, the sodium extract is treated with silver nitrate. Only in the case of fluorine, the silver fluoride formed is soluble, unlike the others which is insoluble. Thus, the precipitate will not be formed and so, this method cannot be used for fluorine detection.

4. Potassium can replace sodium in lassaigne’s test.
a) True
b) False
Answer: a
Clarification: Potassium, like sodium is electropositive in nature. In lassaigne’s test, the elements present in the compound are converted from their covalent form to their ionic form by fusing the compound with sodium metal. Since, potassium has similar characteristics of electro positivity as sodium and since potassium is highly reactive, it can be used instead of sodium in lassaigne’s test.

5. What is Lassaigne’s test extract called as?
a) Fusion extract
b) Sodium fusion extract
c) Lassaigne extract
d) Sodium extract
Answer: b
Clarification: Lassaigne’s test extract is called as sodium fusion extract. The cyanides, sulphides and halides of sodium will be formed. These will be extracted from the fused mass by boiling it with distilled water. Hence, the name of the extract is sodium fusion extract.

6. In the test for nitrogen, the sodium fusion extract is acidified with which of the following?
a) Dilute sulphuric acid
b) Dilute hydrochloric acid
c) Concentrated hydrochloric acid
d) Concentrated sulphuric acid
Answer: d
Clarification: In the test for nitrogen, sodium cyanide first reacts with iron (III) sulphate and forms sodium hexacyanoferrate (II). On heating with concentrated sulphuric acid, some iron (II) ions are oxidized to iron (III)ions which react with sodium hexacyanoferrate (II) to produce iron (III) hexacyanoferrate (II), which is Prussian blue in color. Moreover, nitrogen atoms are soluble in concentrated sulphuric acid.

7. What is the color of the precipitate obtained in the test for sulphur?
a) White
b) Black
c) Violent
d) Blue
Answer: b
Clarification: In the test for sulphur, the sodium fusion extract is acidified with acetic acid and lead acetate is added to it. Once this reaction takes place, a black precipitate is formed. This black precipitate is lead acetate, indicating the presence of sulphur.

8. In case of both nitrogen and sulphur existence, Prussian blue is still the color of the end product.
a) True
b) False
Answer: b
Clarification: In case, both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed. The color formed is blood red and not Prussian blue. This is because, in this case, there are no free cyanide ions.
Na + C + N+ S → NaSCN
Fe³⁺ + SCN^– → [Fe (SCN)] ²⁺ (blood red)

9. A X color precipitate, which is Y in ammonium hydroxide indicates presence of chlorine. Identify X and Y.
a) X = yellowish, Y = soluble
b) X = yellow, Y = insoluble
c) X = white, Y = insoluble
d) X = white, Y = soluble
Answer: d
Clarification: During the detection of chlorine, when the organic compound reacts with sodium, it forms sodium chloride. This sodium chloride gives the white precipitate of silver nitrate with silver nitrate solution. This white precipitate is also soluble in ammonium hydroxide.

10. Which is the oxidizing agent used in the test for phosphorous?
a) Hydrogen peroxide
b) Potassium nitrate
c) Sodium peroxide
d) Nitric acid
Answer: c
Clarification: The phosphorous present in the organic compound is oxidized to phosphate by the oxidizing agent sodium peroxide. This solution is then boiled with nitric acid and treated with ammonium molybdate. As a result, a yellow precipitate is formed, indicating the presence of phosphorous.

Chemistry Questions and Answers for Aptitude test on “Uncertainty in Measurement”.

1. Write 6354000000 in scientific notation.
a) 6.354 x 10⁹
b) 6354 x 10⁶
c) 0.64 x 10¹⁰
d) 6354000 x 10³
Answer: a
Clarification: Scientific notation should have the least possible significant figures as the zeroes are made as to the power of ten. Or simply the power of ten in scientific notation is equal to the number of times the decimal point moved to produce a number between 1 and 10.

2. _________ is referred to as the closeness of different measurements for the same quantity.
a) Accuracy
b) Precision
c) Analysis
d) Dimension
Answer: b
Clarification: Precision is referred to as the closeness of different measurements for the same quantity. Accuracy is the degree to which it’s taken as standard or almost equivalent to it.

3. How many seconds are there in a half day?
a) 86,400 seconds
b) 43,200 seconds
c) 172,800 seconds
d) 3660 seconds
Answer: b
Clarification: Half day means 12 hours. One hour means 60 seconds. And each minute is of 60 seconds. In the above question, we need the total number of seconds so 12 x 60 x 60 seconds = 43,200 seconds.

4. A piece of iron is 5 inches long. How much would it be in centimeters?
a) 12.7 cm
b) 6.35 cm
c) 5 cm
d) 500 cm
Answer: a
Clarification: 1 inch = 2.54 cm : 1 cm = 0.3931 inch. As we now know, how much is 1inch in centimeters, then we need to multiply 5 to 2.54 cm in order to convert it into centimeters. 5 x 2.54 cm = 12.7 cm.

5. How many significant figures does 0.057 have?
a) 2
b) 4
c) 3
d) 0
Answer: a
Clarification: The non-zero digits and any zeros between them are all significant. Leading zeros are not significant. Counting all the significant digits gives us 2.

6. How many significant figures does 63180 have?
a) 5
b) 4
c) 1
d) 2
Answer: b
Clarification: The non-zero digits and any zeros between them are all significant. Since there is no decimal, zeros are not significant. Therefore it’s 4.

7. The exact value is 150m. A students record it as 140.1m in 1st turn and 140.8m in the 2^nd turn. Comment his/her recordings.
a) precise
b) accurate
c) neither precise nor accurate
d) both precise and accurate
Answer: a
Clarification: These values are precise as they are close to each other but are not accurate. Precision is referred to as the closeness of different measurements for the same quantity. Accuracy is the degree to which it’s taken as standard or almost equivalent to it.

8. The exact value is 150m. A students record it as 149.1m in 1st turn and 150.8m in the 2^nd turn. Comment his/her recordings.
a) precise
b) accurate
c) neither precise nor accurate
d) both precise and accurate
Answer: b
Clarification: These values are accurate as they are nearer to the exact value. Precision is referred to as the closeness of different measurements for the same quantity. Accuracy is the degree to which it’s taken as standard or almost equivalent to it.

9. Multiply 1.2 and 3.91. Obtain the result as per the rules of significant figures.
a) 4.692
b) 4.69
c) 5
d) 4.7
Answer: d
Clarification: 1.2 x 3.91 = 4.692. Since 1.2 has two significant figures, the result should not have more than two significant figures, thus, it is 4.7. As per the rules of significant figures, the resultant answer should not have more significant figures the number with a less significant figured number.

10. How many significant figures are there in 60.6?
a) 4
b) 2
c) 3
d) 1
Answer: c
Clarification: The non-zero digits and any zeros between them are all significant. Leading zeros are not significant. Counting all the significant digits gives us 3.