Category: Class 11 Chemistry Objective Questions

Chemistry online quiz on “States of Matter – Gas Laws”.

1. At a constant temperature, the pressure of a gas is given as one atmospheric pressure and 5 liters. When the atmospheric pressure is increased to 2 atm, then what is the volume of the gas?
a) 1 liter
b) 5 liters
c) 10 liters
d) 2.5 liters
Answer: d
Clarification: As we know that, Boyle’s law states at a constant temperature, the pressure of a gas is inversely proportional to its volume so P₁V₁ equals to P₂V₂ by substituting P₁ as one atmospheric pressure V₁ as 5 liters P₁ as to atmospheric pressure we get V₂ as 5/2 that is 2.5 liters.

2. What is the shape of the graph that is drawn between pressure and volume?
a) A straight line
b) Circular
c) Parabola
d) Hyperbola
Answer: d
Clarification: Boyle’s law states that at constant temperature pressure is inversely proportional to the volume of gas so here the graph between the pressure as y-axis volume as x-axis is in the shape of a hyperbola.

3. What is the name of the graph that is drawn, when the temperature is kept constant?
a) Isotherm
b) Isochoric and isobar
c) Isochoric
d) Isobar
Answer: a
Clarification: The graphs with constant temperature plot are isotherms. For example, the graph that is used to detect the Boyle’s law, that is between pressure and volume is an isotherm as the temperature is constant in this graph.

4. There is a ball that will burst if the pressure exceeds 0.12 bars. The pressure of the gas is 1 bar and the volume is 2.5 liters. What can be the maximum volume that the ball can be expanded?
a) 0.12 liters
b) 2.5 liters
c) 0.3 liters
d) 1 liter
Answer: c
Clarification: According to Boyle’s law at a constant temperature, the pressure is inversely proportional to the temperature so here P₁V₁ is equaled to P₁V₂ by equating P₁V₁ is equaled to 1 x 2.5 = 2.5, so the maximum volume of the ball that can be expanded is 2.5/0.12 =0.3 liters.

5. How much does the volume of the gas increase if we increase the temperature by 1 Degree?
a) 273 liters
b) 1 by 273^rd of the original volume of the gas
c) 1 liter
d) Hundred liters
Answer: b
Clarification: According to Charles law, the volume of the fixed gas at constant pressure is directly proportional to the Absolute Temperature of the gas. So we thereby represent this as V_t = V₀(1 + t/273) where V_t is the volume of the gas at temperature t and V₀ is the volume of the gas at 0 degrees Celsius that is absolute temperature.

6. There is a balloon filled with a gas at 26-degree centigrade and has a volume of about 2 liters when the balloon is taken to a place which is at 39-degree centigrade, what would be the volume of the gas that is inside the balloon?
a) 2 liters
b) 3 liters
c) 1.5 liters
d) 0.67 liters
Answer: b
Clarification: As we know that temperature is directly proportional to the volume at constant pressure, 26/39 = 2/ X; so here by equating X equals to 3 liters. Hence required a volume of the balloon at 39 degrees is 3 liters.

7. By observing the below-given figure which of the options do you think is the correct one?

a) P₁ is greater than P₂
b) P₂ is greater than P₁
c) P₁ is equal to P₂
d) P₁ may be equal to P₂
Answer: a
Clarification: By using Boyle’s law, draw a parallel line to volume axis, so as to maintain a constant temperature. Draw perpendicular lines to the point of intersection of pressure lines to constant temperature and volume axis. Now see to the lower the volume, higher the pressure. So P₁ is greater than P₂.

8. An ideal gas of 10 moles occupies _________ volume.
a) 22.4 liters
b) 2.24 liters
c) 224 liters
d) 2240
Answer: c
Clarification: As we know that the number of moles is proportional to the volume as per Avogadro’s law and we also know that an ideal gas at STP occupies 22.4 liters of volume. So here 10 moles of gas occupies 224 liters of volume.

9. When a graph is drawn between the pressure and temperature of the gas it is known as _________
a) isochoric
b) isobar
c) isotherm
d) isotopic
Answer: a
Clarification: When a graph is plotted between pressure on y-axis and temperature on x-axis straight line is formed at a constant volume and this graph is known as isochoric. As we know that gay lussac’s law proposes that at the constant volume the pressure and temperature are directly proportional.

10. At 22 degree Celsius a gas consists of pressure 1.1 bars then what is the temperature when the gas consists a pressure of 2.2 bars?
a) 11 degree Celsius
b) 44 degree Celsius
c) 33 degree Celsius
d) 22 degree Celsius
Answer: b
Clarification: According to Gay-Lussac’s law, at the constant volume, the pressure is directly proportional to the temperature of gas so P₁/P₂ = T₁/T₂ that is 1.1 Bar/2.2 bar = 22 degrees Celsius/44 degree Celsius. So the temperature required is 44 degrees Celsius.

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Chemistry Multiple Choice Questions on “Thermodynamics – Measurement of ∆U and ∆H: Calorimetry”.

1. Bomb calorimeter is __________ in nature.
a) isothermal
b) isochoric
c) isobaric
d) absolute
Answer: b
Clarification: ABomb calorimeter works at constant volume, so it is isochoric in nature. Here, the heat energy that’s measured is only the internal energy. The work done is zero because of no change in volume.

2. If an exothermic reaction occurs in a Bomb calorimeter then the temperature of the water bath ________
a) increases
b) decreases
c) remains constant
d) cannot predict
Answer: a
Clarification: In a Bomb calorimeter, the reaction occurs in a vessel and that is surrounded by the water bath. If it is an exothermic reaction the temperature rises and if it is an endothermic reaction the temperature decreases. Temperature can be measured using a thermometer.

3. Heat capacity of a Bomb calorimeter is given by _______
a) C_V
b) C_P
c) C_M
d) C_B
Answer: a
Clarification: A bomb calorimeter operates at a constant volume i.e. it is an isochoric process. So the heat capacity is C_V which represents heat capacity at constant volume. While C_P represents heat capacity at constant pressure.

4. Bomb calorimeter is used to determine ____________
a) molar heat capacity
b) heat of combustion
c) rate kinetics
d) affinity
Answer: b
Clarification: A bomb calorimeter is used to measure the heat of combustion of a reaction. It has to withstand a large amount of pressure, in order to determine the heat of combustion. It is an isochoric process and the heat energy is equal to the internal energy.

5. “c” the specific heat capacity of a substance is given by temperature difference is given by ΔT and the heat energy is given by Q then what is the mass of the substance?
a) Q/cΔT
b) cΔT /Q
c) QcΔT
d) QΔT
Answer: a
Clarification: Heat energy of a substance is denoted by Q and is given by the expression mcΔT, where m is a mass of the substance, c is a specific heat capacity and ΔT is temperature difference, so the mass of the substance is Q/cΔT.

6. During the process of conversion of ice into the water, the specific heat capacity is given by _______
a) 0
b) positive
c) infinity
d) negative
Answer: c
Clarification: During the phase change of a substance the temperature change is zero. As we know that specific heat capacity = Q/cΔT which is zero; specific heat capacity becomes infinity, so during the process of conversion of ice into the water, the specific heat capacity is infinity.

7. What is the value of specific heat capacity in the adiabatic process?
a) 0
b) infinity
c) positive
d) negative
Answer: a
Clarification: During an adiabatic process the change in total energy is zero, As we know that the specific heat capacity is given by the total heat required by mass X change in temperature, heat energy is zero the specific heat capacity becomes zero.

8. When 1 kg of water at 373 k, is converted into steam how much amount of heat energy is required?
a) 22600 KJ
b) 226 KJ
c) 2260 KJ
d) 22.6 KJ
Answer: c
Clarification: The latent heat of vaporization of water is 2260 KJ/Kg. The heat that is required to convert water at 373 k to steam is given by Q = mL, where m = mass of the water and L = latent heat vaporization of water; heat energy required = 1 kg x 2260 KJ/Kg = 2260 KJ.

9. The total heat energy utilized for increasing the temperature by 4 degrees Kelvin in a 3 kgs substance is 100 KJ what is the specific heat capacity of that substance?
a) 8.34 KJ/g-k
b) 8.34 KJ/Kg-k
c) 8.34 KJKg-k
d) 8.34 KJ/Kg
Answer: b
Clarification: The formula of heat energy is given by the expression: Q = mcΔT, m is the mass of the substance, c is the specific heat capacity and ΔT is the temperature difference. c = Q/mΔT; c = 100KJ/3kg(4k) = 8.34 KJ/Kg-k.

10. The enthalpy, internal energy during a process and change in volume are 500 units, 400 units, and 2 units. What is the pressure that is exerted on the gas during this process?
a) 20 units
b) 80 units
c) 100 units
d) 50 units
Answer: d
Clarification: We know that ΔH = ΔU + PΔV; ΔH is the enthalpy, ΔU is the internal energy, ΔV is the change in volume and P is the pressure. So by substituting the enthalpy, internal energy during a process and change in volume as 500 units, 400 units and 2 units, we get pressure as 50 units.

Chemistry Online Test for Class 11 on “Solubility Equilibria of Sparingly Soluble Salts”.

1.The degree of dissociation of Ammonium hydroxide increases in the presence of Ammonium Chloride because of ______________
a) solubility product
b) common Ion effect
c) hydrolysis of the salt
d) mixed salts
Answer: c
Clarification: Common Ion effect is defined as the separation of the dissociation of a weak electrolyte by the addition of a strong electrolyte having some common ion. Therefore the degree of dissociation of Ammonium hydroxide decreases in the presence of Ammonium Chloride due to common Ion effect.

2. Common Ion effect can be used in which of the following cases?
a) cloth making
b) alcohol purification
c) quantitative analysis
d) qualitative analysis
Answer: d
Clarification: Common Ion effect is used in the purification of common salt, salting out of soap and qualitative analysis. In qualitative analysis, Ammonium Hydroxide is added in the presence of Ammonium Chloride to avoid the precipitation of V group radicals.

3. Hydroxide Ion concentration in calcium hydroxide and barium Hydroxide is an example of ______________ solution.
a) isochoric solution
b) isohydric solutions
c) hypo solution
d) hyper solution
Answer: b
Clarification: In the solution of two electrolytes, if the common ions’ concentration (Hydroxide Ion concentration in calcium hydroxide and barium hydroxide solution) is equal, then on mixing there is zero change in the degree of association in both of the electrolytes, such solutions are called isohydric solutions.

4. For the dissociation of an electrolyte A_xB_y, K_sp is given as [A^y+]^x[B^x+]^y. What is K_sp?
a) solubility product
b) soluble product
c) solution product
d) solvent product
Answer: a
Clarification: Solubility product K_sp is defined as the product of the concentrations of the ions of the salt in its standard solution at a given temperature raised to the power of the ions produced by the dissociation of 1 mole of the salt.

5. Precipitate is formed if ionic product is ______________
a) greater than the solubility product
b) less than the solubility product
c) equal to the solubility product
d) independent of the solubility product
Answer: a
Clarification: The concept of solubility product helps in predicting the formation of the precipitate. In general, if the ionic product is greater than the solubility product, the precipitate is formed and if the ionic product is less than the solubility product, the precipitate is not formed.

6. Solubility product can be used in predicting the solubility of a sparingly soluble salt.
a) true
b) false
Answer: a
Clarification: Yes, we can predict the solubility of a sparingly soluble salt, for example, consider the reaction; A_xB_y = xA^y+ + yB^x+, the solubility of a sparingly soluble salt is given by x^x.y^y.s^x+y, knowing the values of K_sp, x and y, the solubility of the salt can be calculated.

7. A salt is soluble is the solubility is ______________
a) less than 0.01 M
b) in between 0.01 M and 0.1 M
c) greater than 0.01 M
d) greater than 0.1 M
Answer: d
Clarification: A salt is soluble if the solubility is greater than 0.1 M. A salt is slightly soluble if the solubility is between 0.01 M and 0.1 M and the salt is sparingly soluble if the solubility is less than 0.01 M.

8. If K_sp of a salt A₂B₃ is given by 1 x 10^-25. Then find the solubility of the salt?
a) 10^-3
b) 10^-4
c) 10^-5
d) 10^-8
Answer: c
Clarification: For the salt A_xB_y, (A_xB_y = xA^y+ + yB^x+), the solubility of a sparingly soluble salt is given by x^x.y^y.s^x+y. K_sp = x^x.y^y.s^x+y, where x = 2 and y = 3; K_sp = 108S5 = 1 x 10^-25. S = 10^-5. The solubility of the salt is given by 10^-5.

9. The solubility for the salts of the type AB₃ is given by _______________
a) (K_sp 27)^1/4
b) (K_sp/27)^1/5
c) (K_sp/27)^3/4
d) (K_sp/27)^1/4
Answer: d
Clarification: For the salt A_xB_y, (A_xB_y = xA^y+ + yB^x+), the solubility of a sparingly soluble salt is given by x^x.y^y.s^x+y. K_sp = x^x.y^y.s^x+y, where x = 1 and y = 3; K_sp = 27S⁴, by rearranging, we get solubility denoted by S as (K_sp/27)^1/4.

10. Both the solubility product and ionic product are applicable to all types of solutions.
a) true
b) false
Answer: b
Clarification: Solubility of the product is only applicable to the saturated solutions, whereas an ionic product is applicable to all types of solutions. It is because the formation of a precipitate is dependent on the solubility product.

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Chemistry MCQs for NEET Exam on “s-Block Elements – Anomalous Behaviour of Beryllium”.

1. Which of the following element is different from the other group 2 elements?
a) magnesium
b) beryllium
c) calcium
d) strontium
Answer: b
Clarification: Beryllium, differs from the rest of the members of its group due to the following reasons: beryllium has a small atomic and ionic size, it has no vacant d-orbitals and it has a higher charge density comparatively.

2. Beryllium has ______________ melting point and is ______________ than the other members of its family.
a) low, smoother
b) low, harder
c) higher, harder
d) higher, smoother
Answer: c
Clarification: The points of difference are that the beryllium is denser and harder than other members of its family and it has a higher melting point that is 1551 Kelvin while that of magnesium is just 924 Kelvin. So we can say that beryllium has a higher melting point and is harder than other members of its family.

3. Does beryllium react with water?
a) Yes
b) No
c) Maybe
d) Cannot say
Answer: b
Clarification: Beryllium does not react with water even at a higher temperature while the other members of the family liberate hydrogen by reacting with water at room temperature, this is also one of the reasons of the anomalous behaviour of beryllium.

4. Beryllium Hydroxide is ______________ in nature.
a) acidic
b) basic
c) amphoteric
d) cannot say
Answer: c
Clarification: Beryllium oxide (BeO) and beryllium hydroxide [Be(OH)₂] are amphoteric in character they also dissolve in acid to form a salt and beryllate in alkali. Beryllate is any salt containing an anion BeO₂^2-.

5. Beryllium forms beryllium carbide on heating.
a) true
b) false
Answer: a
Clarification: Beryllium when heated with carbon forms beryllium carbide which on reaction with water gives methane, while the other members of the group form ionic carbide which is acetylide which on reaction with water evolves acetylene.

6. Beryllium and Aluminium are ______________ in nature.
a) electropositive
b) electronegative
c) metallic
d) fluorescent
Answer: b
Clarification: Beryllium’s and aluminium action with water is the same, they do not decompose with water or in the presence of water even at hundred degrees centigrade. This is probably due to their less electropositive character that is they are electronegative in nature.

7. Which of the following is a correct formula for sodium meta aluminate?
a) Na₂AlO₄
b) Na₂Al₂O
c) Na₂Al₂O₄
d) NaAl₂O₄
Answer: c
Clarification: When aluminium reacts with alkali, that is when one mole of aluminium reacts with two moles of sodium hydroxide in presence of two moles of water two moles of sodium meta aluminate is formed along with three moles of hydrogen. The chemical formula for sodium meta aluminate is Na₂Al₂O₄.

8. Aluminium and beryllium are ______________ in nature.
a) metallic
b) active
c) passive
d) electropositive
Answer: c
Clarification: Aluminium and beryllium are rendered passive on reaction with concentrated nitric acid, due to the formation of the oxide layer on their surfaces, this is one of the similarities between aluminium and beryllium.

9. Does beryllium form complexes?
a) Yes
b) No
c) Maybe
d) Cannot say
Answer: a
Clarification: Beryllium and aluminium form a number of complexes both form fluoro complex and ions, this is one of the similar chemical properties between beryllium and aluminium as they share a diagonal relationship.

10. Which of the following Statements do you think is true?
a) aluminium and beryllium are not stable in air
b) beryllium and aluminium share a diagonal relationship
c) the ionisation potential of beryllium is lower compared with its group
d) beryllium has a greater oxidation potential
Answer: b
Clarification: The correct statement is a beryllium and aluminium share a diagonal relationship, as aluminium and beryllium are stable in air, the ionization potential of beryllium has higher comparatively with its group and it has a lower oxidation potential.

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Chemistry Problems for NEET Exam on “Organic Chemistry – Quantitative Analysis”.

1. Carbon and hydrogen are detected by heating the compound with ____________ during quantitative analysis.
a) carbon dioxide
b) copper (II) oxide
c) magnesium oxide
d) sulphur dioxide
Answer: b
Clarification: During quantitative analysis, a known quantity of the organic compound is burned in the presence of excess of oxygen and copper (II) oxide. The carbon present in the compound is oxidized to carbon dioxide and the hydrogen is transformed to water.

2. On complete combustion, 0.500 g of an organic compound gave 0.150 of carbon dioxide. Determine the percentage composition of carbon in the compound.
a) 8.18%
b) 81.8%
c) 0.81%
d) 81%
Answer: a
Clarification: The required equation is:
Percentage of carbon = 12 * m2 * 100/ 44 * m
In this case, m = 0.500
m2 = 0.150
Substituting in the equation → Percentage of carbon = 12 * 0.150 * 100/ 44 * 0.500
= 8.18%
Therefore, the percentage of carbon present in the organic compound is 8.18%.

3. The amounts of water can be detected by the increase in the mass of __________ during quantitative analysis of hydrogen.
a) potassium hydroxide
b) sodium hydroxide
c) sodium chloride
d) calcium chloride
Answer: d
Clarification: In the quantitative analysis of hydrogen, hydrogen is converted to water. Then, water is passed through a u-tube containing calcium chloride. Calcium chloride absorbs water and thereby, increases in mass. This increase in the mass of calcium chloride gives the amounts of water present in the organic compound.

4. Dumas method is a method of estimation of nitrogen.
a) True
b) False
Answer: a
Clarification: One of the methods of quantitative analysis of nitrogen includes Dumas method. In Dumas method, the nitrogen containing organic compound is heated with copper oxide in an atmosphere of carbon dioxide and this leads to the formation of free nitrogen in addition to carbon dioxide and water.

5. 0.350 g of an organic compound gave 100 ml of nitrogen collected at 250 K temperature and 700 mm pressure. Calculate the percentage composition of nitrogen in the compound. Use Dumas method.
a) 25.70%
b) 35%
c) 35.71%
d) 36.88%
Answer: c
Clarification: Here, p = 700 mm; v = 100 ml; t =250 K
Volume of nitrogen at STP = p * v * 273/ 760 * t
= 700 * 100 * 273/ 760 * 250
= 100.57 ml
22,400 ml of nitrogen at STP weighs = 28 g
Then, 100.57 ml of nitrogen weighs = 28 * 100.57/22400
= 0.125 g
Therefore, percentage of nitrogen = 0.125 * 100/0.350
= 35.71%
So, percentage of nitrogen in the organic compound is 35.71%.

6. In Kjedahl’s method of estimation of nitrogen, the compound containing nitrogen is heated with __________
a) concentrated hydrochloric acid
b) concentrated sulphuric acid
c) nitric acid
d) dilute hydrochloric acid
Answer: b
Clarification: The compound containing nitrogen is heated with concentrated sulphuric acid. This leads to the conversion of nitrogen to ammonium sulphate. The resulting acid is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction.

7. Using Carius method, find out the percentage of bromine in the compound if 0.450 g of organic compound gave 0.200 g of AgBr. (Given: molar mass of Ag = 108; molar mass of Br = 80)
a) 18%
b) 17%
c) 17.68%
d) 18.91%
Answer: d
Clarification: The molar mass of AgBr = 108 + 80
= 188 g
So, 188 g of AgBr contains 80 g of Br.
Then, 0.200 g of AgBr contains 80 * 0.200/ 188 = 0.085 g bromine
Therefore, percentage of bromine = 0.085 * 100/ 0.450
= 18.91%
Thus, the percentage of bromine (halogen) using Carius method is 18.91%.

8. In the estimation of sulphur, the organic compound is heated with ___________ in a carius tube.
a) sodium hydroxide
b) sodium peroxide
c) potassium hydroxide
d) calcium chloride
Answer: b
Clarification: In the estimation of sulphur, a known quantity of the organic compound is heated with sodium peroxide. This facilitates the sulphur present to be oxidized to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The percentage of sulphur can be estimated from the mass of barium sulphate.

9. Estimation of oxygen can be determined by Carius method.
a) True
b) False
Answer: b
Clarification: Carius method is used for the estimation of halogens present in an organic compound. Percentage of oxygen present in organic compound can be calculated by the difference between the total percentage composition (100) and the sum of the percentages of all other elements. The percentage of oxygen can also be derived from the amount of iodine produced.

10. The elements present in a compound are determined by apparatus called ____________
a) analyzer
b) CHN elemental analyzer
c) chemical analyzer
d) elemental analyzer
Answer: b
Clarification: The elements like carbon, hydrogen, and nitrogen present in an organic compound are determined by an apparatus called as CHN elemental analyzer. The analyzer requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time.

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