NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry
Do you know that single drop of water contain more molecules than there are stars in the entire universe and why compounds always combine in a fixed proportion. Some basic concepts of chemistry is like the ABCs of Chemistry where you get to know what matter is made of, how atoms and molecules interact, and why every tiny calculation in chemistry counts. There are so many examples in our daily lives where basic chemistry plays an important role.
This Story also Contains
- NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 1)
- NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 2)
- NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Short Answer Type
- NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Matching Type
- NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Assertion and Reason Type
- NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Long Answer Type
- Class 11 Chemistry NCERT Chapter 1: Higher Order Thinking Skills (HOTS) Questions
- Approaches to Solve Questions of Chapter 1 Some Basic Concepts of Chemistry
- Main Topics in NCERT Exemplar Class 11 Chemistry Chapter 1
- NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 - Chapter-Wise
- NCERT Solutions for Class 11 Chemistry Chapter-wise
- NCERT Exemplar Class 11 Solutions
- NCERT Solution subject-wise
- NCERT Notes subject-wise
- NCERT Books and NCERT Syllabus
NCERT Exemplar Solutions for Class 11 Chemistry provides a detailed explanation of basic concepts and laws of chemistry and the principles and theories that govern their behavior. This chapter also deals with fundamental ideas related to atoms, moles, and molecules, which are the core of chemistry. Our subject experts design the NCERT Exemplar Solutions to offer a systematic and structured approach to these important concepts and help students to develop a clear understanding of critical concepts by the series of solved examples and conceptual explanations, these solutions provide a valuable resource to enhance performance in board exams as well as in the competitive exams. Students can also check NCERT Solutions to all questions chapter-wise.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 1)
At first, the MCQ questions are covered in the Class 11 Chemistry NCERT Exemplar Solutions Chapter 1 to enhance your knowledge. The concepts are explained in detail in notes available on our website.
|
Student
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Reading
| |
|
|
(i)
|
(ii)
|
|
A
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3.01
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2.99
|
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B
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3.05
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2.95
|
(i) Results of both the students are neither accurate nor precise.
(ii) The results of student A are both precise and accurate.
(iii) Results of student B are neither precise nor accurate.
(iv) The results of student B are both precise and accurate.
Answer:
The answer is option (ii) Results of student A are both precise and accurate.
Explanation: First, we calculate, the average of both the readings of student A and B. Student A's average is 3.00, and student B's average is also 3.00. The correct reading of the mass is 3.00g. Therefore, both the students have gotten an average value close to the correct value and are accurate. But student A's values differ only by 0.02 and are precise while B's readings are not close to each other and differ by 0.1, and thus are not precise.
Therefore, A has readings that are both precise and accurate.
Question 2. A measured temperature on the Fahrenheit scale is 200 °F. What will this reading be on the Celsius scale?
(i) 40 °C
(ii) 94 °C
(iii) 93.3 °C
(iv) 30 °C
Answer:
Explanation:
$^{o}F=\frac{9}{5} (^{o}C)+32$
$^{o}C=\frac{5}{9} (^{o}F-32)$
$^{o}C=\frac{5}{9} (200-32)=93.3$
Question 3. What will be the molarity of a solution which contains 5.85 g of NaCl(s) per 500 mL?
(i) 4 mol L-1
(ii) 20 mol L-1
(iii) 0.2 mol L-1
(iv) 2 mol L-1
Answer:
Explanation: First, calculating the number of moles of NaCl $=\frac{5.85}{58.5}=0.1\; mol/L$
$500\; ml=\frac{500}{1000}=0.5L$
Molarity (M) = number of moles of solute/ Volume of solution in liters
Molarity $=\frac{0.1}{0.5}$
Molarity = 0.2 M
Question 4. If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
(i) 1.5 M
(ii) 1.66 M
(iii) 0.017 M
(iv) 1.59 M
Answer:
Explanation: Using M1V1 = M2V2
$5 \times 500=M_{2} \times 1500$
$M2=\frac{(5 \times 500)}{1500}$
$M_{2}=1.66 M$
Question 5. The number of atoms present in one mole of an element is equal to Avogadro's number. Which of the following elements contains the greatest number of atoms?
(i) 4 g of He
(ii) 46g Na
(iii) 0.40 g of Ca
(iv) 12g He
Answer:
Explanation:
Using the previous formula,
12 g He=3 moles Of He
Number of atoms of He$=3\times N_{A}$
Number of atoms of He$=3\times\left ( 6.022\times 10^{23} \right )$
Question 6. If the concentration of glucose $(C_{6}H_{12}O_{6})$ in the blood is 0.9 g L-1, what will be the molarity of glucose in the blood?
(i) 5 M
(ii) 50 M
(iii) 0.005 M
(iv) 0.5 M
Answer:
The answer is option (iii) 0.005 MExplanation:
Given; Concentration of glucose- $(C_{6}H_{12}O_{6})$ in blood =0.9/L
Molar mass of glucose = 180 g/mol
Concentration/molar mass $=\frac{0.9}{180}$
$=\frac{1}{200}\; moles=0.005\; moles$
Therefore, the Molarity of glucose in blood is 0.005 M.
Question 7. What will be the molality of the solution containing 18.25 g of HCL gas in 500 g of water?
(i) 0.1 m
(ii) 1M
(iii) 0.5 m
(iv) 1m
Answer:
The answer is option (iv) 1mExplanation:
Number. of moles of HCl $=\frac{18.25}{36.5}=0.5$
Molality (m) = Number of moles (in the solute)/ Mass of the solute (kg)
Molality $=\frac{0.5\times 1000}{500}$
Molality = 1m
Question 8. One mole of any Substance contains $6.022\; \; 10^{23}$ of Atoms/molecules of $H_{2}SO_{4}$ present in the 100 mL of 0.02M $H_{2}SO_{4}$ solution is
(i) $12.044 \times 10^{20}$ molecules
(ii) $6.022 \times 10^{23}$ molecules
(iii) $1 \times 10^{23}$molecules
(iv) $12.044 \times 10^{23}$ molecules
Answer:
Explanation:
Given: Molarity = 0.02 M
Given; Volume of solution = 100ml = 0.1 L
Number of moles Of $H_{2}SO_{4}$ = Molarity × Volume ( litres)
Number of molecules of $H_{2}SO_{4}$ = $2 \times 10^{-3} \times 6.022 \times 10^{23}$
Number of molecules = $12.044\times 10^{20}$ Molecules
Question 9. What is the mass percent of carbon in carbon dioxide?
(i) 0.034%
(ii) 27.27%
(iii) 3.4%
(iv) 28.7%
Answer:
The answer is option (ii) 27.27%
Explanation:Molar mass of $CO_{2}$ $= (1\times 12) + (2 \times 16) = 44 g$
Therefore, 1 g of $CO_{2}$ contains 1 g of carbon
44 g of $CO_{2}$ therefore contains 12 g of carbon.
% of C in $CO_{2}$ =$=\frac{12}{44}\times 100=27.27\; ^{o}/_{o}$
Question 10. The empirical formula and molecular mass of a compound are $CH_{2}O$ and 180 g respectively. What will be the molecular formula of the compound?
(i) $C_{9}H_{18}O_{9}$
(ii) $CH_{2}O$
(iii) $C_{6}H_{12}O_{6}$
(iv) $C_{2}H_{4}O_{2}$
Answer:
The answer is the option (iii) $C_{6}H_{12}O_{6}$Explanation:
Empirical formula of $CH_{2}O$ is $CH_{2}O$
Calculating empirical formula mass = $12+(1\times 2)+16=30g$
Given: Molecular mass= 180g
n = Molecular mass/ empirical formula mass
$n=\frac{180}{30}=6$
Therefore, Molecular formula = $n\times CH_{2}O=6\times CH_{2}O=C_{6}H_{12}O_{6}$
Question 11. If the density of a solution is 3.12 g mL-1, the mass of 1.5 mL solution in significant figures is _______.
(i) 4.7g
(ii) 4680 × 10-3 g
(iii) 4.680g
(iv) 46.80g
Answer:
Explanation:
Given, the density of a solution = 3.12 g mL-1
Volume = 1.5 mL
Mass = Volume × Density
= 1.5mL × 3.12g m/L
= 4.68 g, which when rounded to 2 significant figures is 4.7 g.
Question 12. Which of the following statements about a compound is incorrect?
(i) A molecule of a compound has atoms of different elements.
(ii) A compound cannot be separated into its constituent elements by physical methods of separation.
(iii) A compound retains the physical properties of its constituent elements.
(iv) The ratio of atoms of different elements in a compound is fixed.
Answer:
The answer is option (iii). A compound retains the physical properties of its constituent elements.Explanation:
The compound can be completely indifferent, in contrast to the physical properties of its constituent elements.
Question 13. Which of the following statements is correct about the reaction given below:
$4Fe(s) + 3O_{2} (g) \rightarrow 2Fe_{2}O_{3} (g)$
(i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows Law of conservation of mass.
(ii) Total mass of reactants = total mass of product; therefore, the Law of multiple proportions is followed.
(iii) The amount of $Fe_{2}O_{3}$ can be increased by taking any one of the reactants (iron or oxygen) in excess.
(iv) The amount of $Fe_{2}O_{3}$ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.
Answer:
Explanation:
Keeping the Law of conservation in mind, we evaluate both sides of every equation. The mass of the reactants should be equal to the mass of the product, and this is only followed in equation (i), therefore it is correct.
Question 14. Which of the following reactions is not correct according to the law of conservation of mass?
(i)$2Mg(s) + O_{2} (g) \rightarrow 2MgO(s)$
(ii) $C_{2} H_{8} (g) + O_{2} (g) \rightarrow CO_{2} (g) + H_{2}O(g)$
(iii) $P_{4}(s) + 5O_{2}(g) \rightarrow P_{4}O_{10}(s)$
(iv) $CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)$
Answer:
The answer is the option (ii) $C_{2} H_{8} (g) + O_{2} (g) \rightarrow CO_{2} (g) + H_{2}O(g)$
Explanation:
Again, following the Law of energy, the number of atoms on the reactant side should be equal to the product side. Option (ii) does not follow the rule.
Question 15. Which of the following statements indicates that the Law of Multiple Proportions is being followed?
(i) A sample of carbon dioxide taken from any source will always have carbon and oxygen in the simple ratio of 1:2,
(ii) Carbon forms two oxides namely $CO_{2}$ and $CO$. where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.
(iii) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.
(iv) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour.
Answer:
Explanation:
According to the Law of Multiple Proportions,
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 2)
The class 11 chemistry chapter 1 Some Basic Concepts of Chemistry questions are provided here with simple explanations. Learn more through these advanced MCQs
Question 16. One mole of oxygen gas at STP is equal to ________
(i) $6.022\times 10^{23}$ molecules of oxygen
(ii) $6.022\times 10^{23}$ atoms of oxygen
(iii) 16 g of oxygen
(iv) 32 g of oxygen
Answer:
The answer is the options (i) and (iv)Explanation:
1 mole of oxygen gas at STP = $6.022\times 10^{23}$ molecules of oxygen (which is the Avogadro number)
1 mole $O_{2}=32g \; of \; O_{2}$
Thus, one mole of $O_{2}$ is equal to the molecular weight of oxygen and the Avogadro number.
Question 17. Sulphuric acid reacts with sodium hydroxide as follows :
$H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$
When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0. 1M sodium hydroxide solution, the amount of sodium sulfate and its molarity in the solution obtained is
(i) 0.1 mol L-1
(ii) 7.10 g
(iii) 0.025 mol L-1
(iv) 3.55 g
Answer:
$H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$
Given, (0.1 M), Moles of H2SO4 = 0.1 moles
Given, (0.1M), Moles of NaOH = 0.1 moles
The ratio of H2SO4 and NaOH is 1:2, Using the ratio, we figure for 0.1 mole of H2SO4 , 0.2 moles of NaOH is used, which is not available.
Therefore if 0.1 moles of NaOH reacts, then only $0.05(0.1/2)$ moles of H2SO4 is used. NaOH is the limiting reactant, and the reaction and products are evaluated according to the limiting reactant.
Therefore the number of moles formed of Na2SO4 will be 0.05 as well.
This is because 2 moles of NaOH produces 1 mole of H2SO4, therefore 1 mole of NaOH will logically produce 0.05 moles of Na2SO4.
Therefore, mass of Na2SO4 = 0.05 × (molar mass)
0.05 × (23 ×2 + 32 + 16× 4)
0.05 × 142 = 7.1 g
Molarity = Number of moles / Volume in L
Molarity = 0.05/ 2 L = 0.023 mol/L
Question 18. Which of the following pairs have the same number of atoms?
(i) 16 g of $O_{2}$ (g) and 4 g of $H_{2}$ (g)
(ii) 16 g of $O_{2}$ and 44 g of $CO_{2}$
(iii) 28 g of $N_{2}$ and 32 g of $O_{2}$
(iv) 12 g of C (s) and 23 g of Na(s)
Answer:
The answer is the options (iii) and (iv)Explanation:
One mole is measured as the amount of a substance that equal number of atoms as contained in exactly 12 g of Carbon.
1 mole of C = 12g C
While 23 grams of Na (mass) = 1 mole of Na
28g N2 = 1 mole of nitrogen = 2 × atoms of nitrogen
This is $2 \times (6.022 \times 10^{23})$atoms of nitrogen
32g O2 = 1 mole of oxygen = $2 \times (6.022 \times 10^{23})$ atoms of oxygen
Question 19. Which of the following solutions have the same concentration?
(i) 20 g of NaOH in 200 mL of solution
(ii) 0.5 mole of KCl in 200 mL of solution
(iii) 40 g of NaOH in 100 mL of solution
(iv) 20 g Of KOH in 200 mL of solution
Answer:
The answer is the options (i) and (ii)Explanation: (i) 20 g NaOH in 200 mL solution.
$\frac{20}{40}=0.5$ mol of NaOH
(ii) Molar concentration Of NaOH
$\frac{0.5\; mol}{0.2\; L}=2.5 \; M$ as 200 mL = 0.2 L
Molar concentration of KCl = $\frac{0.5\; mol}{0.2\; L}=2.5 \; M$
Question 20. 16 g of oxygen has the same molecules as in
(i) 16 g of CO
(ii) 28 g of $N_{2}$
(iii) 14 g of $N_{2}$
(iv)) 1.0 g of $H_{2}$
Answer:
Explanation: 1 mole or the number of molecules of oxygen in 16g Of oxygen is
$\frac{16}{32}\times \left ( 6.023 \times 10^{23} \right )$
$=0.5\times \left ( 6.023\times 10^{23} \right )$ molecules
In 14g or N2 and 1 g of H2 contain 0.5 moles. Therefore they will also contain $0.5\times \left ( 6.023\times 10^{23} \right )$ molecules
Number of molecules of N2
$=\frac{14}{28}\times \left ( 6.023\times 10^{23} \right )$
$=0.5\times \left ( 6.023\times 10^{23} \right )$ molecules
Number of molecules of H2
$=\frac{1}{2}\times 6.023\times 10^{23}$
$=0.5\times \left ( 6.023\times 10^{23} \right )$ molecules
Question 21. Which of the following terms are unitless?
(i) Molality
(ii) Molarity
(iii) Mole fraction
(iv) Mass percent
Answer:
The answer is the options (iii) and (iv)Explanation:
Mole Fraction: It is the particular ratio of the substance in the solution in comparison to the total components of the solution. It is the ratio of the number of moles of the particular component to the total number of moles in the solution.
While, Mass percent=$\frac{Mass\; of\; solute}{Mass\; of\; solution}\times 100$
Question 22. One of the Statements of Dalton's atomic theory is given below; "Compounds are formed when atoms of different elements combine in a fixed ratio". Which of the following laws is not related to this statement?
(i) Law of conservation of mass
(ii) Law Of definite proportions
(iii) Law of multiple proportions
(iv) Avogadro's Law
Answer:
The answer is the options (i) and (iv)Explanation:
For (iv), Avogadro suggested that equal volumes of gases will contain ean qual number of molecules if the temperature and pressure remain the same.
While the Law of conservation of mass states that matter can neither be created nor destroyed but just transformed from one form to another.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Short Answer Type
Here some short answer type questions from NCERT Exemplar Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry are given for practice. This section contains important questions that are asked in the exams.Practice short answer types from the questions below.
Question 23. What will be the mass of one atom of C-12 in grams?
Answer:
Therefore, mass of one atom of $C=\frac{12}{6.023\times 10^{23}}$ , this is $1.992648\times 10^{-23}$ g
Answer:
$\frac{2.5\times 1.25\times 3.5}{2.01}$ has 2 significant figures.
Question 25. What is the symbol for the SI unit of a mole? How is the mole defined?
Answer:
Question 26. What is the difference between molality and molarity?
Answer:
Molarity (M) = $\frac{Number\; of\; moles\; of\; solute}{Volume\; of \; solution\; in\; litres}$
Molality: It is the number of moles of solute present in 1 kg of solvent. It is denoted by m. It is independent of temperature.
Thus, Molality (m)= $\frac{Number\; of\; moles\; of\; the\; solute}{Mass\; of\; Solvent\; in\; kg}$
Question 27. Calculate the mass percent of calcium, phosphorus, and oxygen in calcium Phosphate; $Ca_{3} (PO_{4}��)_{2}$
Answer:
$=\left ( 3\times 40 \right )+\left [ 2\left ( 31+16\times 4 \right ) \right ]=310$
Mass percent of Calcium =$\frac{3x \; Atomic \; mass \; of \; calcium}{Total\; molecular \; mass\; of \; calcium \; Phosphate}\times 100$
$=\frac{3\times 40}{310}\times 100= 38.71\; ^{o}/_{o}$
Likewise,
Mass percent of phosphorus = $\frac{2x \; Atomic \; mass \; of \; Phosphorous}{Total\; molecular \; mass\; of \; calcium\; Phosphate}\times 100$
$=\frac{2\times 31}{310}\times 100= 20\; ^{o}/_{o}$
Mass percent of Oxygen=$\frac{2x \; Atomic \; mass \; of \; Oxygen}{Total\; molecular \; mass\; of \; calcium\; Phosphate}\times 100$
$=\frac{2\times 64}{310}\times 100= 41.29\; ^{o}/_{o}$
Answer:
The Law being obeyed in this experiment is Gay Lussac's Law of Gaseous Volumes. The Law states, that gases react with each other in a simple ratio by volume, provided the gases are maintained at a constant temperature and pressure.Answer:
(a) Yes, the statement is true.(b) The statement is true due to the Law of multiple proportions.
(c) Citing an example; carbon combines with oxygen to form ; CO and CO
The mass of oxygen, when reacted with a fixed mass of carbon (12 g) has a simple ratio i.e.,16:32 = 1:2
Question 30. Calculate the average atomic mass of hydrogen using the following data:
|
Isotope |
% Natural Abundance |
Molar Mass |
|
1H |
99.985 |
1 |
|
2H |
0.015 |
2 |
Answer:
Average atomic mass = $\frac{(Natural\; Abudance\; of\; ^{1}H \times Molar\; mass)+(Natural\; Abudance\; of\; ^{2}H \times Atomic\; mass)}{100}$$=\frac{(99.985\times 1)+(0.015\times 2)}{100}$
$=\frac{100.015}{100}=1.00015\; \upsilon$
Given that 65.3 g of Zn reacts with HCI to form 22.7 of H2(g)
Therefore, 32.65 g of Zn at STP reacts with HCI to form $=\frac{22.7\times 32.65}{65.3}=11.35L$
Therefore, 32.65 g of Zn at STP reacts with HCI to form $=\frac{22.7\times 32.65}{65.3}=11.35L$
Question 32. The density of 3 molal solutions of NaOH is 1.110 g mL-1. Calculate the molarity of the solution.
Answer:
Therefore, the Mass of Solution-= Mass of SoIvent + Mass of Solute
$=1000g+(3\times 40\; g)$
$=1120\; g$
Volume of Solution =$\frac{MASS}{DENSITY}$
$=\frac{1120}{1.110\; mL}=100.90\; mL$
Given the density of solution =1.110 g m/L-1
Since 1009 mL solution has 3 mols Of NaOH
Molarity (M) = $\frac{Number \; of\; moles\; of\; solute}{Volume\; of\; solution\; in\; litres}$
$\frac{3\; mol}{1009.00}=2.97\; M$
Answer:
No, it will not. The molality of any solution is indifferent to temperature changes as molality is expressed in mass and mass is unaffected by temperature fluctuations.Molality (m) = $\frac{number\; of\; moles\; of\; the\; solute}{Mass\; of\; Solvent\; in\; kg}$
As per the information given in the question, the Mass of NaOH = 4g
Therefore, Number of moles of NaOH=$\frac{4}{40}=0.1\; mol$
Number of moles of water $=\frac{36}{18}=2\; mols$
Mole fraction of water =$\frac{Number\; of\; moles\; of\; water}{Total\; number\; of\; moles(water+NaOH)}\frac{2}{2.1}=0.952$
Mole fraction of NaOH = $\frac{Number\; of\; moles\; of\; NaOH}{Total\; number\; of\; moles(water+NaOH)}\frac{0.1}{2.1}=0.047$
Now, mass of solution = Mass of water + Mass of NaOH=$36+4=40 \; g$
Volume $=\frac{40}{1}=40\; mL$
Molarity in solution = $\frac{Number\; of\; moles\; of\; solute}{Volume\; of\; solute\; in\; Litre}=\frac{0.1\; mol\; NaOH}{0.04L}=2.5 M$
Therefore, Number of moles of NaOH=$\frac{4}{40}=0.1\; mol$
Number of moles of water $=\frac{36}{18}=2\; mols$
Mole fraction of water =$\frac{Number\; of\; moles\; of\; water}{Total\; number\; of\; moles(water+NaOH)}\frac{2}{2.1}=0.952$
Mole fraction of NaOH = $\frac{Number\; of\; moles\; of\; NaOH}{Total\; number\; of\; moles(water+NaOH)}\frac{0.1}{2.1}=0.047$
Now, mass of solution = Mass of water + Mass of NaOH=$36+4=40 \; g$
Volume $=\frac{40}{1}=40\; mL$
Molarity in solution = $\frac{Number\; of\; moles\; of\; solute}{Volume\; of\; solute\; in\; Litre}=\frac{0.1\; mol\; NaOH}{0.04L}=2.5 M$
Answer:
$2A + 4B \rightarrow 3C + 4D$A and B are in the ratio- 2:4, 2 mols of 'A' require 4 mols of 'B' .
(i) For 5 mols of A, 10 mols of 'B' will be required using the ratio of 2:4.
i.e. 5 mol of $A\times \frac{4\; mol\; of\; B}{2\; mol\; of\; A}=$10 mol of
But only 6 mols of B is available therefore it is the limiting reagent
(ii), Amount of 'C' formed is depends on 'B' ,
4 mols of 'B' give 3 mols of C, the ratio is 4:3.
Therefore, 6 mols of B will give, 6 mol of $B\times \frac{3\; mol\; of\; C}{4\; mol\; of\; C}=4.5\; moles\; of\; C$
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Matching Type
Class 11 chemistry chapter 1 Some Basic Concepts of Chemistry important questions are discussed below. These are generally asked in exams to test your knowledge. These exemplar solutions is quite helpful for competitive exams.
Question 36. Match the following:
|
Column I
|
Column II
|
|
A. 88 g of CO2
|
a. 0.25 mol
|
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B. 6.022 x 1023 molecules of H2O
|
b. 2 mol
|
|
C. 5.6 litres of O2 at STP
|
c. 1 mol
|
|
D. 96 g of O2
|
d. 6.022 x 1023 molecules
|
|
E. 1 mole of any gas
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e. 3 mol
|
Answer:
i) – b
ii) – c
iii) – a
iv) - e
v) – d
Question 37. Match the following physical quantities with units
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Column I (Physical)
|
Column II (Unit)
|
|
(i) Molarity
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a. g mL-1
|
|
(ii) Mole fraction
|
b. mol
|
|
(iii) Mole
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c. Pascal
|
|
(iv) Molality
|
d. Unitless
|
|
(v) Pressure
|
e. mol L-1
|
|
(vi) Luminous intensity
|
f. Candela
|
|
(vii) Density.
|
g. mol kg-1
|
|
(viii) Mass
|
h. N m-1
|
|
|
i. kg
|
Answer:
(i) – e(ii) – d
(iii) – b
(iv) – g
(v) – c
(vi) – f
(vii) – a
(viii) - i
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Assertion and Reason Type
This is one of the most important sections covered in the NCERT exemplar solutions Class 11 chemistry chapter 1. These questions will improve your critical thinking.The most typical and important section for exams
Question 38. In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): The empirical mass of ethene is half of its molecular mass.
Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) A is true but R is false.
(iii) A is false but R is true.
(iv) Both A and R are false.
Answer:
Explanation: Molecular formula = n × Empirical formula mass
Empirical formula of ethene = $CH_{2}$
Empirical formula mass of ethene = 14 amu
$=\frac{1}{2}\times mass\; of\; ethene$
The empirical formula shows that ethene has (C: H)= 1:2
Question 39. In the following questions ,a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): One atomic mass unit is defined as one-twelfth of the mass of one carbon- 12 atoms.
Reason (R): Carbon-12 isotope is the most abundant isotope of carbon and has been Chosen as standard.
(i) Both A and R are true, and R is the correct explanation Of A.
(ii) Both A and R are true, but R is not the correct explanation of A
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:
The answer is option (ii) Both A and R are true, but R is not the correct explanation of A.Explanation: Carbon has isotopes and Carbon-12 is one of the isotopes represented as 12C.
12C has a mass of exactly 12 amu –(atomic mass units) and the mass of all other elements is calculated with respect to this. One amu is the mass exactly of one-twelfth the mass of one 12C atom.
Question 40. In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Significant figures for 0.200 is 3 whereas for 200 it is 1.
Reason (R): A zero at the end or right of a number is significant provided they are not on the right side of the decimal point.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not a correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:
Explanation:
Zero's at the end of a number which does not have a decimal point, depends on the measurement.
Question 41. In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Combustion Of 16 g Of methane, gives 18 g Of water.
Reason (R): In the combustion of methane. water is one of the products.
(i) Both A and R are true but R is not the correct explanation Of A.
(ii) A is true but R is false.
(iii) A is false but R is true.
(iv) Both A and R are false.
Answer:
Explanation:
$CH_{4} (g) + 2O_{2} (g) \rightarrow CO_{2} �(g) + 2H_{2}O (g)$
16g of CH4 on complete combustion will give 36g of water.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Long Answer Type
The following are the long-answer type questions that needs more practice. These are the chemistry chapter 1 Some Basic Concepts of Chemistry important questions that are asked in the exams.
Answer:
$P_{1}=1 \; atm$$T_{1}=273\; K$
We will be calculating V1 now,
We know that 32 g of oxygen occupies 22.4 L of volume at STP+
Hence, 1.6 g Of O2 will occupy,
$1.6\; g\; of\; Oxygen\times \frac{22.4\; L}{32\; g\; of\; oxygen}=1.12\; L$
$Hence\; V_{1}=1.12\; L$
We know that, $p_{2}=p_{1}/2=\frac{1}{2}=0.5 \; atm$
Now we must calculate $V_{2}$
As per Boyle's law, $P_{1}V_{1}=P_{2}V_{2}$
Therefore, $V_{2}=\frac{P_{1}V_{1}}{2}$
$\frac{1\; atm\times 1.12\; L}{2}=0.5\; atm\times v_{2}$
Therefore, $V_{2}=2.24L$
(iii) Number of molecules of O2 that are present in the vessel,
$= 6.023\times 10^{23}\times \frac{1.6}{32}=3.011\times 10^{22}$
Answer:
Number of moles of HCI=$250\; mL\times \frac{0.76\; M}{1000}=0.19\; mol$Given: Mass of $CaCO_{3}$ = 1000 g
Number of moles of $CaCO_{3}$ = $\frac{1000\; g}{100\; g}=10\; \; mol$
Referring to the equation, 1 mol of $CaCO_{3}$ (s) needs 2 mol of HCI (aq), the ratio is 1:2.
Hence, for 10 mol of $CaCO_{3}$ (s), the number of moles of HCI required would be:
$10\; Mol\; CaCO_{3}\times \frac{2\; Mol\; HCl}{1\; Mol\; CaCO_{3}}=20\; Mol\; HCl (aq)$
But only 0.19 mol of HCL (aq) is available and therefore, HCL is a limiting reagent.
The amount of $CaCl_{2}$ will depend on the amount of HCl present. 2 mol of HCl (aq), gets 1 mol of $CaCl_{2}$, and the ratio is 2:1.
Thus, 0.19 mol of HCl will get=$0.19\; Mol\; HCl\times \frac{1\; mol\; CaCl_{2}}{2\; mol\; HCl}=0.095\; mol\; of\; CaCl_{2}$
0.095 ×molar mass of $CaCl_{2}$= 0.095 ×111 = 10.54 g
Answer:
The Law of multiple proportions states that "If two elements can combine to form more than one compound, the mass of element combines with the fixed mass of the other element and is in the ratio of small whole numbers." For example, hydrogen reacts with oxygen to form water andhydrogen peroxide.
Hydrogen + Oxygen → Water
2g 16g 18g
Hydrogen + Oxygen → Hydrogen Peroxide
2g 32g 34g
Here, the masses of oxygen combine with a fixed mass of hydrogen and have a simple ratio of 16:32 or 1:2. This Law reflects that reactants always combine in fixed proportions.
Answer:
Applying the Law of multiple proportionsFor the combination, AB
1 g of A combines with $\frac{5}{2}g\; of\; B=2.5g\; of\; B$
For AB2
1 g of A combines$\frac{10}{2}g\; of\; B=5g\; of\; B$
For A2B
1 g of A combines with$\frac{5}{4}g\; of\; B=1.25g\; of\; B$
For A2B2
1 g of A combines with$\frac{15}{4}g\; of\; B=3.75g\; of\; B$
Thus, it is proved that the Law of multiple proportions is applicable.
Class 11 Chemistry NCERT Chapter 1: Higher Order Thinking Skills (HOTS) Questions
Some Class 11 chemistry chapter 1 some basic concepts of chemistry questions and answers are given below that will help you tackle complex problems. The questions below will help you evaluate your understanding of the concepts.
Question 1 : Among $10^{-9} \mathrm{~g}$ (each) of the following elements, which one will have the highest number of atoms?
Element : $\mathrm{Pb}, \mathrm{Po}, \mathrm{Pr}$ and Pt
Options
i) Pb
ii)Po
iii)Pr
iv)Pt
Solution:
$\text { No. of atoms }=\frac{\text { Mass}}{\text { MolarMas }(\mathrm{g} / \mathrm{mol})} \times \mathrm{N}_{\mathrm{A}}$
From this formula, it's clear that for a given mass, the element with the smallest molar mass will have the greatest number of atoms, as it appears in the denominator.
$\mathrm{M}_{\mathrm{P}_{\mathrm{o}}}=209$
$\mathrm{M}_{\mathrm{pr}}=141$
$\mathrm{M}_{\mathrm{Pb}}=207$
$\mathrm{M}_{\mathrm{Pt}}=195$
Pr has the lowest molar mass ( $M_{\mathrm{Pr}}=141 \mathrm{~g} / \mathrm{mol}$ ), $10^{-9}$ grams of Pr will contain the highest number of atoms compared to the other elements listed.
Hence, the correct answer is option (2).
Question 2: $\mathrm{CaCO}_3(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{CO}_2(\mathrm{~g}) \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
Consider the above reaction, what mass of $\mathrm{CaCl}_2$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of $\mathrm{CaCO}_3$ ?
(Given : Molar mass of $\mathrm{Ca}, \mathrm{C}, \mathrm{O}, \mathrm{H}$ and Cl are 40 , $12,16,1$ and $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$, respectively)
Options
i) 3.908 g
ii) 2.636 g
iii) 10.545 g
iv) 5.272 g
Solution:
$\mathrm{CaCO}_3(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{CaCl}_2(a q)+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)$
$
\text { Volume }=250 \mathrm{~mL}=0.250 \mathrm{~L}
$
Moles of $\mathrm{HCl}=M \times V=0.76 \times 0.250=0.19 \mathrm{~mol}$
From the equation:
2 mol HCl reacts with $1 \mathrm{~mol} \mathrm{CaCO}_3$
Moles of $\mathrm{CaCO}_3=\frac{0.19}{2}=0.095 \mathrm{~mol}$
Molar mass of $\mathrm{CaCO}_3=100 \mathrm{~g} / \mathrm{mol}$
Available moles of $\mathrm{CaCO}_3=\frac{1000}{100}=10 \mathrm{~mol}$
From the equation:
$1 \mathrm{~mol} \mathrm{CaCl}_2$ is formed per 2 mol HCl
$
\text { Moles of } \mathrm{CaCl}_2=\frac{0.19}{2}=0.095 \mathrm{~mol}
$
Molar mass of $\mathrm{CaCl}_2=40(\mathrm{Ca})+2 \times 35.5(\mathrm{Cl})=111 \mathrm{~g} / \mathrm{mol}$
$
\text { Mass }=0.095 \times 111=10.545 \mathrm{~g}
$
Hence, the correct answer is option (3).
Question 3. On combustion 0.210 g of an organic compound containing $\mathrm{C}, \mathrm{H}$ and O gave $0.127 \mathrm{~g} \mathrm{H}_2 \mathrm{O}$ and $0.307 \mathrm{~g} \mathrm{CO}_2$. The percentages of hydrogen and oxygen in the given organic compound respectively are:
(1) $53.41,39.6$
(2) $6.72,53.41$
(3) $7.55,43.85$
(4) $6.72,39.87$
Solution:
Mass of organic compound $=0.210 \mathrm{~g}$
Mass of water formed $=0.127 \mathrm{~g}$
Mass of $\mathrm{CO}_2$ formed $=0.307 \mathrm{~g}$
Mass of hydrogen $=\frac{0.127 \times 2}{18}=0.014 \mathrm{~g}$
Percentage of hydrogen $=\frac{0.014 \times 100}{0.210}=6.72 \%$
Mass of carbon $=\frac{0.307 \times 12}{44}=0.084 \mathrm{~g}$
Percentage of carbon $=\frac{0.084 \times 100}{0.210}=39.87 \%$
Percentage of oxygen $= 100- (39.87 + 6.72) = 53.41 \%$
Hence, the correct answer is option (2).
Approaches to Solve Questions of Chapter 1 Some Basic Concepts of Chemistry
To solve class 11 chemistry chapter 1 Some Basic Concepts of Chemistry questions, it is important to follow a systematic approach. Given below the approaches to solve these questions effectively.
1.Understand the basic concepts
While solving Some Basic Concepts of Chemistry questions students are guided to first understand the basic concepts like SI Units & Measurements, Significant figures, Scientific notation, Dimensional analysis, and Laws of Chemical Combination. Students can also refer to Class 11 Some Basic Concepts of Chemistry Notes.
2. Practice the formulas of the mole concept like
Students must practice formulas like Atomic Mass and number of moles, Mole & Avogadro’s Number, Percentage Composition like Empirical Formula & Molecular Formula, and Stoichiometry.
3. Calculations and practice
Practice is the key to success; students must practice maximum questions.
- Conversions of mass into moles and moles into the number of particles, empirical & Molecular formulas, finding % composition, converting % to grams, etc
- Find the moles of each element, determine the simplest ratio, which is the empirical formula, and use the particular molecular mass to find the actual formula.
- Stoichiometric Problems like using balanced equations, applying mole ratios, and identifying limiting reagents.
4. Solve Numerical Problems
Given below are some important topics from Class 11 Chemistry Chapter 1 NCERT Exemplar Solution that are asked in exams so students must practice them to score good marks
- Density & Molar Volume for gases, use 1 mole = 22.4 L at STP.
- Molarity (M) = Moles of solute / Volume of solution (L)
- Molality (m) = Moles of solute / Mass of solvent (kg)
- Mole Fraction (χ) = Moles of component / Total moles
- Percentage Purity: (Mass of pure substance / Total mass) × 100.
5. Practice questions
Practice questions from NCERT textbooks, as these questions are asked directly in boards and other competitive exams. For revision students can follow Class 11 Some Basic Concepts of Chemistry Notes. While solving questions, students must take care of a few basic points like, converting the units where required, and significant figures must be thoroughly checked.
Main Topics in NCERT Exemplar Class 11 Chemistry Chapter 1
The following are the important topics of chapter 1. Practice more at NCERT Class 11 chemistry chapter 1 exemplar solutions.
1.1 Importance of Chemistry
1.2 Nature of matter
1.3 Properties of matter and their measurement
1.4 Uncertainty In Measurement
1.5 Laws of Chemical Combinations
1.6 Dalton Atomic Theory
1.7 Atomic and Molecular masses
1.8 Mole Concept and Molar Mass
1.9 Percentage composition
1.10 Stoichiometry and stoichiometric calculation
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 - Chapter-Wise
Follow the links below to get chapter-wise NCERT Class 11 solutions for your exam preparation.
NCERT Solutions for Class 11 Chemistry Chapter-wise
Click on the links below to access the NCERT Class 11 solutions for all the chapters to make your preparations better.
NCERT Exemplar Class 11 Solutions
Excel your preparation with NCERT exemplar solutions. Click on the link below
NCERT Solution subject-wise
The NCERT subject-wise solutions will help you broaden your concepts and will also help in revision. Learn more from Class 11 NCERT notes.
NCERT Notes subject-wise
You can follow the links given in the table below to get access to the Class 11 NCERT notes.
NCERT Books and NCERT Syllabus
You can find links to the Class 11 NCERT chemistry book and syllabus for the respective subjects.
Frequently Asked Questions (FAQs)
Q: What are significant figures? Why are they important?
A:
Significant figures are the meaningful digits in a measured quantity. They indicate the precision of measurements and calculations.
Q: What is the difference between empirical and molecular formulas?
A:
Empirical formula is the simplest whole-number ratio of atoms (e.g., CH₂O for glucose) while molecular formula is the actual number of atoms (e.g., C₆H₁₂O₆ for glucose).
Q: What is the difference between homogenous and heterogeneous mixtures?
A:
A Homogeneous Mixture is a mixture that has a uniform composition throughout the mixture. E.g., saltwater, air, sugar dissolved in water while heterogeneous Mixture is a mixture that does not have a uniform composition. E.g., sand and water, oil and water, a salad.
Q: What is the difference between mass and weight?
A:
Mass is the measure of the amount of matter (constant, SI unit: kg) while weight is a Force exerted by gravity on mass (varies with location, SI unit: N).
Q: What is a limiting reagent?
A:
The reactant that gets consumed first, limiting the amount of product formed in a reaction
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