The chapter contains important topics such as atomic and molecular mass, molar mass, gram atomic mass and gram molecular mass. ANAND CLASSES (A School Of Competitions) provides accurate solutions of Mole Concept MCQs.
JEE Main Mole Concept Previous Year Questions With Solutions
1. Number of atoms in the following samples of substances is the largest in :
(1) 127.0g of iodine
(2) 48.0g of magnesium
(3) 71.0g of chlorine
(4) 4.0g of hydrogen
Solution:
1 mole represents 6.023×1023 particles.
1 mole of iodine atom= 6.023×1023
Given 127.0g of iodine.
no. of iodine atom = 1 mole of iodine
1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg
Given 48g of Mg = 2×6.023×1023
no. of Mg = 2 moles of Mg
1 mole of chlorine atom= 6.023× 1023
no. of chlorine atom = 35.5g of chlorine atom
Given 71g of chlorine atom=2× 6.023× 1023
no. of chlorine atom = 6.023×1023
2 moles of chlorine atom.
Given that 4g of hydrogen atom.
will be equal to 4 × 6.023 × 1023
no. of atoms of hydrogen= 4 moles of hydrogen atom.
Hence option(4) is the answer.
2. The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1. If one molecule of the above compound ( CxHyOZ ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CxHyOz is :
(1) C2H4O
(2) C3H4O2
(3) C2H4O3
(4) C3H6O3
Solution:
Given the ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1.
Atomic mass of carbon = 12
Atomic mass of Hydrogen = 1
If we have x atoms of Carbon and y atoms of Hydrogen,
12*x = 6(1*y)
12x = 6y
So y = 2x
Given one molecule of compound ( CxHyOZ ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O
CXHy + O2 → xCO2+ (y/2)H2O
Put y = 2x in above equation
CXH2x + O2 → xCO2+ xH2O
Oxygen needed = 2x+x = 3x
z is half of oxygen required to burn.
So z = 3x/2 = 1.5 x
Check the given options which satisfies z = 1.5x.
So the empirical formula is C2H4O3.
Hence option (3) is the answer.
3. The concentrated sulphuric acid that is peddled commercially is 95% H2SO4 by weight. If the density of this commercial acid is 1.834 g cm-3, the molarity of this solution is :-
(1) 17.8 M
(2) 15.7 M
(3) 10.5 M
(4) 12.0 M
Solution:
Given Density = 1.834
1 ml solution contains 1.834 g
1000 ml solution will contain 1834 g
95% H2SO4 means 100 gm contain 95 gm H2SO4
Mass of solute = (95/100)×1834
Molecular weight of H2SO4 = 98
Molarity = No. of moles/ volume = mass of solute/98
= (95/100)×(1834/98)
= 17.8 M
Hence option (1) is the answer.
4. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is :
(1) 1 : 8
(2) 3 : 16
(3) 1 : 4
(4) 7 : 32
Solution:
Given ratio of masses of oxygen and nitrogen = 1:4
Let mass of O2 = w
Mass of N2 = 4w
Molecules of O2 = w/(32×NA)
Molecules of N2 = 4w/(28×NA)
Ratio of number of molecules = w/(32×NA)÷4w/(28×NA)
= w/(32×NA)×(28×NA)/4w
= 7/32
So the ratio is 7:32.
Hence option (4) is the answer.
5. 3g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :
(1) 42 mg
(2) 54 mg
(3) 18 mg
(4) 36 mg
Solution:
Molarity of CH3COOH solution = mass of acetic acid/molar mass)/volume of solution in litre
Acetic acid is monobasic.
0.042 = W/(60×0.05)
W = 0.042×60×0.05 = 0.126 g
Amount of acetic acid actually adsorbed = 0.180-0.126 = 54mg
Amount of charcoal available = 3 g
So amount of acetic acid adsorbed per gram of charcoal = 54mg×1g/3.0g = 18 mg
Hence option (3) is the answer.
6. The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is
(1) 2.05 M
(2) 0.50 M
(3) 1.78 M
(4) 1.02 M
Solution:
Given density of solution = 1.15g/mL
mass of solution = 1000+120 = 1120 gm
Molar mass = 60
Volume = mass /density of solution
= 1120/1.15
No. of moles = 120/60 = 2
Molarity = No. of moles/ volume
= 2÷ (1120×10-3/1.15)
= 2×1.15×1000 /1120)
= 2.05 M
Hence option (1) is the answer.
7. The ratio of number of oxygen atoms (O) in 16.0g oxygen (O), 28.0 g carbon monoxide (CO) and 32.0g oxygen (O2) is :
(Atomic mass :C =12, O =16 and Avogadro’s constant NA = 6.0 * 1023 mol-1)
(1) 3 : 1 : 1
(2) 1 : 1 : 2
(3) 3 : 1 : 2
(4) 1 : 1 : 1
Solution:
Molar mass of O3 = 48
Given 16 g O3 . So no. of moles of O3 = 16/48 = ⅓
1 mole = 3 ×NA oxygen atoms
So 1/3 mole = NA×3×1/3 no of atoms
= NA oxygen atoms
Molar mass of CO = 28
Given 28 g CO. So no of moles = 28/28 = 1
No. of atoms = 1×NA = NA
Molar mass of O2 = 32
Given 32g O2
No. of moles = 32/32 = 1
No.of atoms = 1×NA = NA
So the ratio is 1:1:1
Hence option (4) is the answer.
8. When CO2 (g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 litre of CO2 (g) over red hot coke, the total volume of the gases increased to 700 mL. The composition of the gaseous mixture at STP is :
(1) CO2 = 200 mL: CO = 500mL
(2) CO2 = 350 mL: CO = 350mL
(3) CO2 = 0.0 mL: CO = 700mL
(4) CO2 = 300 mL: CO = 400mL
Solution:
CO2 (g) + C (s) → 2CO (g)
Total volume = 700 ml = 0.7 L
0.5+x = 0.7
x = .2L = 200 mL
CO2 (g) = 0.5-0.2 = 300ml
CO (g) = 2x = 400 mL
Hence option (4) is the answer.
9. An open vessel at 300 K is heated till ⅖ th of the air in it is expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated is :
(1) 750 K
(2) 400 K
(3) 500 K
(4) 1500K
Solution:
At constant V and P, n1T1 = n2T2
n1 = n
n2 = n-2n/5 = 3n/5
T1 = 300 K
300 n = (3n/5) T2
T2 = 300×5/3 = 500 K
Hence option (3) is the answer.
10. The density of 3M solution of sodium chloride is 1.252 g mL-1. The molality of the solution will be (molar mass, NaCl = 58.5 g mol-1)
(1) 2.18 m
(2) 3.00 m
(3) 2.60 m
(4) 2.79 m
Solution:
Given Molar mass of NaCl = 58.5 g
M = 3 mol L-1
Mass of weight W2 of NaCl in 1L solution W2 = 3×58.5 = 175.5g
Mass of L solution = V × d
= 1000 ×1.25 = 1250g
Mass of H2O in solution (W1) = 1250-175.5 = 1074g
m = W2×1000/Mw2 ×W1 = (175.5×1000)/58.5×1074.5 = 2.79m
Hence option (4) is the answer.
11. 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine with which of the following HCl solution?
(1) 100 mL of 0.2 N HCl
(2) 400 mL of 0.2 N HCl
(3) 100 mL of 0.1 N HCl
(4) 200 mL of 0.2 N HCl
Solution:
NH2CONH2 + 2NaOH → Na2CO3 + 2NH3
2 mole of urea ≡ one mole of NH3
one mole of NH3 = one mole of HCl
So one mole of HCl = 2 mole of urea = 2×0.6/60 = 0.02 mol.
Hence option (1) is the answer.
12. Calculate the mass of FeSO4.7H2O which must be added in 100 kg of wheat to get 10 PPM of Fe.
Solution:
Ppm = (Mass of Fe/total mass)×106
Total mass = 100 kg = 100 × 1000 g
Mass of Fe = (ppm × total mass )/106
= 10× 100 × 1000/106 = 1 g
Molecular mass of FeSO4.7H2O = 278
Mass of one Fe = 56 g
56 g of Fe → 278 g of FeSO4.7H2O
So 1 g of Fe → 278/56 = 4.96 g
Hence 4.96 g is the answer.
13. Given a solution of HNO3 of density 1.4 g/mL and 63% w/w. Determine the molarity of HNO3 solution.
Solution:
Density = mass/volume of solution
Volume = mass / density = 100g/1.4 g/ml = (100/1.4)ml
Molarity = no. of moles of solute /Volume of solution( l) = 1.4×1000/100 = 14 M
Hence 14 M is the answer.
14. A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be
(1) MCl2
(2) MCl4
(3) MCl5
(4) MCl3
Solution:
Given vapour density = 94.8
Vapour density = molecular mass/2
Molecular mass = 94.8×2 = 189.6
Given 74.75% chlorine.
So 74.75/100 * 189.6 = 141.72 g of chloride is there.
Then the number of atoms of chloride will be 141.72/35.5 =3.97 which is approximately 4.
So the formula of metal chloride will be MCl4.
Hence option (2) is the answer.
15. 10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration?
(1) 0.57 M
(2) 5.7 M
(3) 11.4 M
(4) 1.14 M
Solution:
No. of moles of NaOH in 10 mL of 2 M solution = (10/1000)×2 = 0.02 mol
Number of moles of NaOH in 200 mL of 0.5M solution = (200/1000)×0.5 = 0.1 mol
Total number of moles of NaOH = 0.02+0.1 = 0.12 mol
Total volume = 10+200 = 210 mL = 0.210 L
Final concentration = 0.12/0.210 = 0.57 M
Hence option (1) is the answer.
16. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution?
(1) 0.086
(2) 0.050
(3) 0.100
(4) 0.190
Solution:
We know mole fraction = moles of solute/(moles of solute + moles of solvent)
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CBSE Class 11 Chemistry Syllabus is a vast which needs a clear understanding of the concepts and topics. Knowing CBSE Class 11 Chemistry syllabus helps students to understand the course structure of Chemistry.
Unit-wise CBSE Class 11 Syllabus for Chemistry
Below is a list of detailed information on each unit for Class 11 Students.
UNIT I – Some Basic Concepts of Chemistry
General Introduction: Importance and scope of Chemistry.
Nature of matter, laws of chemical combination, Dalton’s atomic theory: concept of elements, atoms and molecules.
Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.
UNIT II – Structure of Atom
Discovery of Electron, Proton and Neutron, atomic number, isotopes and isobars. Thomson’s model and its limitations. Rutherford’s model and its limitations, Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals.
UNIT III – Classification of Elements and Periodicity in Properties
Significance of classification, brief history of the development of periodic table, modern periodic law and the present form of periodic table, periodic trends in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100.
UNIT IV – Chemical Bonding and Molecular Structure
Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules(qualitative idea only), Hydrogen bond.
UNIT V – Chemical Thermodynamics
Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics – internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction) Introduction of entropy as a state function, Gibb’s energy change for spontaneous and nonspontaneous processes. Third law of thermodynamics (brief introduction).
UNIT VI – Equilibrium
Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle, ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization, ionization of poly basic acids, acid strength, concept of pH, hydrolysis of salts (elementary idea), buffer solution, Henderson Equation, solubility product, common ion effect (with illustrative examples).
UNIT VII – Redox Reactions
Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions.
UNIT VIII – Organic Chemistry: Some basic Principles and Techniques
General introduction, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.
UNIT IX – Hydrocarbons
Classification of Hydrocarbons Aliphatic Hydrocarbons: Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions. Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition. Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of – hydrogen, halogens, hydrogen halides and water.
Aromatic Hydrocarbons:
Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.
To know the CBSE Syllabus for all the classes from 1 to 12, visit the Syllabus page of CBSE. Meanwhile, to get the Practical Syllabus of Class 11 Chemistry, read on to find out more about the syllabus and related information in this page.
CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme
In Chemistry subject, practical also plays a vital role in improving their academic scores in the subject. The overall weightage of Chemistry practical mentioned in the CBSE Class 11 Chemistry syllabus is 30 marks. So, students must try their best to score well in practicals along with theory. It will help in increasing their overall academic score.
CBSE Class 11 Chemistry Practical Syllabus
The experiments will be conducted under the supervision of subject teacher. CBSE Chemistry Practicals is for 30 marks. This contribute to the overall practical marks for the subject.
The table below consists of evaluation scheme of practical exams.
Evaluation Scheme
Marks
Volumetric Analysis
08
Salt Analysis
08
Content Based Experiment
06
Project Work
04
Class record and viva
04
Total
30
CBSE Syllabus for Class 11 Chemistry Practical
Micro-chemical methods are available for several of the practical experiments. Wherever possible such techniques should be used.
A. Basic Laboratory Techniques 1. Cutting glass tube and glass rod 2. Bending a glass tube 3. Drawing out a glass jet 4. Boring a cork
B. Characterization and Purification of Chemical Substances 1. Determination of melting point of an organic compound. 2. Determination of boiling point of an organic compound. 3. Crystallization of impure sample of any one of the following: Alum, Copper Sulphate, Benzoic Acid.
C. Experiments based on pH
1. Any one of the following experiments:
Determination of pH of some solutions obtained from fruit juices, solution of known and varied concentrations of acids, bases and salts using pH paper or universal indicator.
Comparing the pH of solutions of strong and weak acids of same concentration.
Study the pH change in the titration of a strong base using universal indicator.
2. Study the pH change by common-ion in case of weak acids and weak bases.
D. Chemical Equilibrium One of the following experiments:
1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing the concentration of either of the ions. 2. Study the shift in equilibrium between [Co(H2O)6] 2+ and chloride ions by changing the concentration of either of the ions.
E. Quantitative Estimation i. Using a mechanical balance/electronic balance. ii. Preparation of standard solution of Oxalic acid. iii. Determination of strength of a given solution of Sodium hydroxide by titrating it against standard solution of Oxalic acid. iv. Preparation of standard solution of Sodium carbonate. v. Determination of strength of a given solution of hydrochloric acid by titrating it against standard Sodium Carbonatesolution.
F. Qualitative Analysis 1) Determination of one anion and one cation in a given salt Cations‐ Pb2+, Cu2+, As3+, Al3+, Fe3+, Mn2+, Ni2+, Zn2+, Co2+, Ca2+, Sr2+, Ba2+, Mg2+, NH4+ Anions – (CO3)2‐ , S2‐, NO2‐ , SO32‐, SO2‐ , NO ‐ , Cl‐ , Br‐, I‐, PO43‐ , C2O2‐ ,CH3COO‐ (Note: Insoluble salts excluded)
2) Detection of ‐ Nitrogen, Sulphur, Chlorine in organic compounds.
G) PROJECTS Scientific investigations involving laboratory testing and collecting information from other sources.
A few suggested projects are as follows:
Checking the bacterial contamination in drinking water by testing sulphide ion
Study of the methods of purification of water.
Testing the hardness, presence of Iron, Fluoride, Chloride, etc., depending upon the regional variation in drinking water and study of causes of presence of these ions above permissible limit (if any).
Investigation of the foaming capacity of different washing soaps and the effect of addition of Sodium carbonate on it.
Study the acidity of different samples of tea leaves.
Determination of the rate of evaporation of different liquids Study the effect of acids and bases on the tensile strength of fibres.
Study of acidity of fruit and vegetable juices.
Note: Any other investigatory project, which involves about 10 periods of work, can be chosen with theapproval of the teacher.
Practical Examination for Visually Impaired Students of Class 11
Below is a list of practicals for the visually impaired students.
A. List of apparatus for identification for assessment in practicals (All experiments) Beaker, tripod stand, wire gauze, glass rod, funnel, filter paper, Bunsen burner, test tube, test tube stand, dropper, test tube holder, ignition tube, china dish, tongs, standard flask, pipette, burette, conical flask, clamp stand, dropper, wash bottle • Odour detection in qualitative analysis • Procedure/Setup of the apparatus
B. List of Experiments A. Characterization and Purification of Chemical Substances 1. Crystallization of an impure sample of any one of the following: copper sulphate, benzoic acid B. Experiments based on pH 1. Determination of pH of some solutions obtained from fruit juices, solutions of known and varied concentrations of acids, bases and salts using pH paper 2. Comparing the pH of solutions of strong and weak acids of same concentration.
C. Chemical Equilibrium 1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing the concentration of eitherions. 2. Study the shift in equilibrium between [Co(H2O)6]2+ and chloride ions by changing the concentration of either of the ions.
D. Quantitative estimation 1. Preparation of standard solution of oxalic acid. 2. Determination of molarity of a given solution of sodium hydroxide by titrating it against standard solution of oxalic acid.
E. Qualitative Analysis 1. Determination of one anion and one cation in a given salt 2. Cations – NH+4 Anions – (CO3)2-, S2-, (SO3)2-, Cl-, CH3COO- (Note: insoluble salts excluded) 3. Detection of Nitrogen in the given organic compound. 4. Detection of Halogen in the given organic compound.
Note: The above practicals may be carried out in an experiential manner rather than recording observations.
We hope students must have found this information on CBSE Syllabus useful for their studying Chemistry. Learn Maths & Science in interactive and fun loving ways with ANAND CLASSES (A School Of Competitions) App/Tablet.
Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus
Q1
How many units are in the CBSE Class 11 Chemistry Syllabus?
There are 9 units in the CBSE Class 11 Chemistry Syllabus. Students can access various study materials for the chapters mentioned in this article for free at ANAND CLASSES (A School Of Competitions).
Q2
What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?
The total marks for the practicals as per the CBSE Class 11 Chemistry Syllabus is 30. It includes volumetric analysis, content-based experiment, salt analysis, class record, project work and viva.
Q3
Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?
The organic chemistry chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry.
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