Category: Class 11 Chemistry Objective Questions

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.

To practice Chemistry Online Test for Class 11,

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.

To practice Chemistry MCQs for NEET Exam,

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.

To practice Chemistry Problems for NEET Exam,

This set of Chemistry Multiple Choice Questions on “Electronic Configurations and Types of Elements: s, p, d and f Blocks”.

1. Name the element that belongs to s-block but is placed in the p-block.
a) Hydrogen
b) Helium
c) Argon
d) Aluminum
Answer: b
Clarification: Helium’s electronic configuration is 1s². As the last electron is filled in s-orbital, it belongs to s-block. Since the 1^st shell cannot accommodate any orbital than s, 1s² is completely filled orbital, hence making it, a noble gas. Noble gases are placed in p-block.

2. __________ has both the characteristics of Alkali metals and halogens.
a) Helium
b) Chlorine
c) Sodium
d) Hydrogen
Answer: d
Clarification: As per the outer shell configuration of hydrogen (that is 1s¹), it has only one electron in s-orbital making it eligible as an Alkali metal. It requires only 1 electron to obtain a noble gas configuration, which is a characteristic of halogen.

3. ns¹ and ns² are the outer shell configurations of elements in s-block.
a) True
b) False
Answer: a
Clarification: Yes, ns¹ and ns² are the outer shell configurations of elements in s-block. The reason behind this is that they are ready to lose 1 electron or 2 electrons depending on the group number and have low ionization enthalpies.

4. The p-block elements along with s-block elements are called as ________ elements.
a) Inner transition
b) Representative
c) Radioactive
d) Transition
Answer: b
Clarification: The p-block elements comprise of elements from group-13 to group-18 while s-block elements are 1^st and 2nd groups. They two together form “Representative elements” or “Main group elements”.

5. What happens to the non-metallic nature as we move from left to right in groups?
a) Increases
b) Decreases
c) Remains constant
d) Irregular
Answer: a
Clarification: Non-metallic nature is defined as a tendency to gain electrons thus having high negative electron gain enthalpies, but the 18th group has no reactivity. So the non-metallic nature increases from left to right in groups leaving noble gases.

6. The outer shell electronic configuration of transition block elements is given by ________
a) (n-1)d^1-10ns²
b) (n-1)d^1-10(n-1)s^0-2
c) (n-1)d^1-10ns^0-2
d) nd^1-10ns^0-2
Answer: c
Clarification: Transition block elemts are d-block elements. The electrons fill in the d-orbital, and s-orbital orbital gets varied for maintaing stability of atoms. Therefore the outer shell electronic configuration of transition block elements is given by (n-1)d^1-10ns^0-2.

7. Which of the following is not true about transuranium elements?
a) Atomic number > 92
b) Elements after Uranium
c) Decay radioactively as they are unstable
d) Example is Thorium
Answer: d
Clarification: The elements after Uranium (Z = 92) are known as transuranium elements, they are unstable and decay radioactively into other elements. But the atomic number of Thorium is 90, hence it is not a transuranium element.

8. Chalcogens are the elements of Group ___________
a) 18
b) 16
c) 12
d) 2
Answer: b
Clarification: Group 16 consists of elements that belong to chalcogens also known as the oxygen family. The elements of this group are oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and Polonium (Po). They are called so because they are mostly found in earth’s crust.

9. Which of the following is not a Metalloid?
a) Germanium
b) Silicon
c) Aluminum
d) Tellurium
Answer: c
Clarification: Metalloids are those with both the properties of metals and non-metals. They are also known as Semi-metals. They are 8 metalloids known till date, namely boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At).

10. Metals consist of ______% of the elements known.
a) 78
b) 32
c) 22
d) 68
Answer: a
Clarification: Elements are classified into metals, non-metals, and metalloids based on their properties. While 78% of the elements that are known today are metals. Metallic nature is prominent on the left side of the periodic table.

Chemistry Question Paper on “Thermodynamics – Enthalpy Change, ∆rH of a Reaction – Reaction Enthalpy”.

1. The standard state of a substance is considered when the temperature is 298 k and the pressure is ____________
a) 1 ATM
b) 1 bar
c) 1 Pascal
d) 760 mm HG
Answer: b
Clarification: All the standard states of a substance are considered when the temperature is 298 Kelvin and the pressure is 1 bar. 1 bar = 0.987 atmospheric pressure = 10000 Pascal = 750.0617 mm of Mercury.

2. All the enthalpies of fusion are positive.
a) true
b) false
Answer: a
Clarification: Fusion is a process of conversion of liquid to solid the enthalpy is energy that is required for a process. As the melting of a solid is endothermic, the enthalpies of fusion are positive so the above statement is true.

3. Consider that, a ball is immersed in water at room temperature and then taken out having 18 grams of water on it, how much amount of energy is required to dry that water at room temperature?
a) 41.43 KJ/mol
b) 49.53 KJ/mol
c) 41.3 KJ/mol
d) 41.53 KJ/mol
Answer: d
Clarification: Heat required to eliminate water : n x Δ_vapH^– = (1 mol) × (44.01 kJ mol^–1) = 44.01 kJ mol^-1. Δ_vapor = Δ_vapH – ΔnRT = 44.01 kJ mol^-1 – 1×8.314 J/K-mol x 298 k x 10^-3 = 41.53 KJ/mol. So the amount of energy needed is 41.53 KJ/mol.

4. Calculate the internal energy change when 2 moles of water at 0 degrees converts into ice at 0-degree centigrade?
a) 12 KJ per mole
b) 6 KJ per mole
c) 1 KJ per mole
d) 102 KJ per mole
Answer: a
Clarification: Energy change when 1 mol of water at 0-degree centigrade changes into ice at 0 degrees in centigrade is 6 kJ/mol, So the internal energy change when 2 moles of water at 0 degrees converts into ice at 0 degrees is 12 kJ/mole.

5. What is a change in energy if 18 grams of water is heated from room temperature to 20 degrees above it?
a) 1.50 KJ
b) 0.506 KJ
c) 1.06 KJ
d) 1.506 KJ
Answer: d
Clarification: We know that Q = msΔT, where Q is the energy,m is the mass of water, s is the specific heat of water and T is the temperature. So the change in energy required here = 18 g x 4.184 J/g-K x 20K = 1.506KJ.

6. When a chemical reaction is reversed the value of enthalpy is reversed in sign.
a) true
b) false
Answer: a
Clarification: For example, the formation of ammonia has an enthalpy of -91.8 KJ per mole and the decomposition of ammonia has an enthalpy of + 91.8 KJ per Mol. So the above statement that when a chemical reaction is reversed the value of enthalpy is reversed in the sign is true.

7. Consider the equation 2 H₂ + O₂ → 2 H₂O, what does the 2 in the coefficient of H₂O molecule represent?
a) number of particles
b) the number of molecules
c) number of moles
d) number of atoms
Answer: c
Clarification: In a balanced thermochemical equation, the coefficients always refer to the number of the moles (but never molecules) of reactants and products involved in a reaction so 2 in the coefficient of H₂O refers to the number of the moles of water.

8. What is the unit of standard enthalpy of fusion or molar enthalpy of fusion?
a) KJ Mol
b) KJ per Mol
c) Mol per KJ
d) 1/KJ Mol
Answer: b
Clarification: The enthalpy change that occurs during melting of one mole of a solid substance in the standard state is called standard enthalpy of fusion or molar enthalpy of fusion, it is represented by the symbol Δ_fusH^–, the units of this are KJ per Mol.

9. Which of the following is not an application of Hess’s law?
a) determination of heat of formation
b) determination of heat of transition
c) determination of Gibb’s energy
d) determination of heat of hydration
Answer: c
Clarification: The following are the applications of Hess’s law; determination of heat of formation, determination of heat of transition and determination of heat of hydration, also to calculate bond energies.

10. ΔH_r = Σ ΔH_f[products] – Σ ΔH_f[reactants].
a) true
b) false
Answer: a
Clarification: The equation ΔH_r = Σ ΔH_f[products] – Σ ΔH_f[reactants] says that the enthalpy of a reaction is the difference between the enthalpy of products and enthalpy of reactants. The above statement regarding enthalpy is true.

To practice all questions papers on Chemistry,