Chapter 1 - Some Basic Concepts of Chemistry

NCERT Solutions Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 1 (Some Basic Concepts of Chemistry) explain the basic principles of chemistry, including the mole concept, atomic mass, and stoichiometry. This chapter helps students understand how to measure the quantities of substances and calculate the amounts of reactants and products in chemical reactions. These concepts are important for learning other chapters of Chemistry.

NCERT Solutions for Class 11 Chemistry Chapter 1 by ALLEN help students understand each topic through clear explanations and step-by-step numerical solutions. Based on the NCERT syllabus, these solutions help students prepare for CBSE exams and strengthen their understanding of concepts needed for JEE and NEET. From calculating the number of particles in a substance to solving limiting reagent problems, students can practise important numerical questions and improve their understanding of chemical calculations.

1.0Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry: Key Concepts

Class 11 Chemistry Chapter 1, Some Basic Concepts of Chemistry, introduces the basic principles of chemistry and explains how to measure the quantities of substances involved in chemical reactions. The main topics covered in this chapter are:

Nature of Matter: Classification of matter into elements, compounds, and mixtures.
Laws of Chemical Combination: Understanding the laws of conservation of mass, definite proportions, multiple proportions, and Gay-Lussac’s law of gaseous volumes.
Dalton’s Atomic Theory: Learning the main postulates of Dalton’s atomic theory and how they explain the basic nature of matter.
Atomic and Molecular Masses: Understanding atomic mass, molecular mass, average atomic mass, and the atomic mass unit (u).
Mole Concept and Molar Mass: Learning the relationship between moles, mass, and Avogadro’s constant ((6.022 \times 10^{23})).
Percentage Composition: Calculating the mass percentage of each element in a compound and finding its empirical and molecular formulas.
Stoichiometry: Balancing chemical equations and calculating the quantities of reactants and products involved in a chemical reaction.
Limiting Reagent: Identifying the reactant that gets consumed first and determines the maximum amount of product formed.
Concentration Terms: Understanding how to express the concentration of a solution using mass percentage, mole fraction, molarity (M), and molality (m).

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 1

NCERT EXERCISE

Calculate the molecular mass of the following: (i) $H_{2} O$ (ii) $CO_{2}$ (iii) $CH_{4}$

Ans.

(i) $H_{2} O$ : The molecular mass of water, $H_{2} O = (2 \times$ Atomic mass of hydrogen $) +$ ( $1 \times$ Atomic mass of oxygen $= [2 (1.0084) + 1 (16.00 u)]$ $= 2.016 u + 16.00 u = 18.016 = 18.02 u$ (ii) $CO_{2}$ : The molecular mass of carbon dioxide, $CO_{2} = (1 \times$ Atomic mass of carbon $) +$ ( $2 \times$ Atomic mass of oxygen)

= [1 (12.011 u) + 2 (16.00 u)] = 12.011 u + 32.00 u = 44.01 u

(iii) $CH_{4}$ : The molecular mass of methane, $CH_{4} = (1 \times$ Atomic mass of carbon $) +$ (4 × Atomic mass of hydrogen)

Calculate the mass percent of different elements present in sodium sulphate $(Na_{2} SO_{4})$ . Ans. The molecular formula of sodium sulphate is $Na_{2} SO_{4}$ . Molar mass of $Na_{2} SO_{4}$

= [(2 \times 23.0) + (32.066) + 4 (16.00)] = 142.066 g

Mass percent of an element

= \frac{Mass of that element in the compound}{Molar mass of the compound} \times 100

∴ Mass percent of sodium :

= \frac{46.0 g}{142.066 g} \times 100 = 32.379 = 32.4%

Mass percent of sulphur:

= \frac{13.066 g}{142.066 g} \times 100 = 22.57 = 22.6%

Mass percent of oxygen:

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. Ans. % of iron by mass = 69.9 % [Given] % of oxygen by mass $= 30.1%$ [Given] Relative moles of iron in iron oxide:

= \frac{% of iron by mass}{Atomic mass of iron} = \frac{69.9}{55.85} = 1.25

Relative moles of oxygen in iron oxide:

= \frac{% of oxygen by mass}{Atomic mass of oxygen} = \frac{30.1}{16.00} = 1.88

Simplest molar ratio of iron to oxygen:

= 1.25 : 1.88 = 1 : 1.5 ≃ 2 : 3

∴ The empirical formula of the iron oxide is $Fe_{2} O_{3}$ .

Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. Ans. The balanced reaction of combustion of carbon can be written as:

1mole C_{(s)} + lyole (32 g) O_{(g)} \to limole (44 g) CO_{2 (g)}

(i) As per the balanced equation, 1 mole of carbon burns in1 mole of dioxygen (air) to produce1 mole of carbon dioxide. (ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant. (iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

Calculate the mass of sodium acetate ( $CH_{3} COONa$ ) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is $82.0245 g mol^{- 1}$ . Ans. 0.375 M aqueous solution of sodium acetate $\equiv 1000 mL$ of solution containing 0.375 moles of sodium acetate ∴ Number of moles of sodium acetate in 500 mL $= \frac{0.375}{1000} \times 500 = 0.1875$ mole Molar mass of sodium acetate $= 82.0245 g mole^{- 1}$ (Given) ∴ Required mass of sodium acetate $= (82.0245 g mol^{- 1}) (0.1875 mole) = 15.38 g$
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, $1.41 g mL^{- 1}$ and the mass per cent of nitric acid in it being 69%. Ans. Mass percent of nitric acid in the sample $= 69%$ [Given] Thus, 100 g of nitric acid contains 69 g of nitric acid by mass. Molar mass of nitric acid $(HNO_{3})$

= {1 + 14 + 3 (16)} g mol^{- 1} = 1 + 14 + 48 = 63 g mol^{- 1} = 63 g mol^{- 1}

∴ Number of moles in 69 g of $HNO_{3}$

= \frac{69 g}{63 g mol ^{- 1}} = 1.095 mol

Volume of 100g of nitric acid solution Concentration of nitric acid

= \frac{Mass of solution}{density of solution} = \frac{100 g}{1.41 mL ^{- 1}} = 70.92 mL \equiv 70.92 \times 1 0^{- 3} L

Concentration of nitric acid $= \frac{1.095 mole}{70.92 \times 1 0 ^{- 3} L}$ $= 15.44 mol / L$ ∴ Concentration of nitric acid $= 15.44 mol / L$

How much copper can be obtained from 100 g of copper sulphate $(CuSO_{4})$ ? Ans. 1 mole of $CuSO_{4}$ contains 1 mole of copper. Molar mass of $CuSO_{4}$

= (63.5) + (32.00) + 4 (16.00) = 63.5 + 32.00 + 64.00 = 159.5 g

159.5 g of $CuSO_{4}$ contains 63.5 of copper. $\Rightarrow 100 g$ of $CuSO_{4}$ will contain $= \frac{63.5 \times 100 g}{159.5}$ of copper. Amount of copper that can be obtained from

Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is $159.69 g mol^{- 1}$ . Ans. Mass percent of iron $(Fe) = 69.9%$ (Given) Mass percent of oxygen $(O) = 30.1%$ (Given) Number of moles of iron present in the

oxide = \frac{69.90}{55.85} = 1.25

Number of moles of oxygen present in the

oxide = \frac{30.1}{16.0} = 1.88

Ratio of iron to oxygen in the oxide,

= 1.5 : 1.88 = \frac{1.25}{1.25} : \frac{1.88}{1.25} = 1 : 1.5 = 2 : 3

The empirical formula of the oxide is $Fe_{2} O_{3}$ . Empirical formula mass of $Fe_{2} O_{3}$

= [2 (55.85) + 3 (16.00)] g

Molar mass of $Fe_{2} O_{3} = 159.69 g$

∴ n = \frac{Molar mass}{Emprical formul mass} = \frac{159.69 g}{159.7 g} = 0.999 = 1 (approx)

Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the empirical formula of the given oxide is $Fe_{2} O_{3}$ and n is 1. Hence, the molecular formula of the oxide is $Fe_{2} O_{3}$ .

Calculate the atomic mass (average) of chlorine using the following data:

	% Natural Abundance	Molar Mass
³⁵Cl	75.77	34.9689
³⁵Cl	24.23	36.9659

Ans. The average atomic mass of chlorine

= [(of ^{3} Cl Fractionalabundance) (of ^{33} Cl Molar mass) + (of ^{3} Cl Fractionalabundance) (of ^{37} Cl Molar mass)] = [{\frac{75.77}{100} (34.9689 u)} + {\frac{24.23}{100}} (36.9659 u)] = 26.4959 + 8.9568 = 35.4527 u

∴ The average atomic mass of chlorine

In three moles of ethane $(C_{2} H_{6})$ , calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane.

Ans.

(i) 1 mole of $C_{2} H_{6}$ contains 2 moles of carbon atoms. ∴ Number of moles of carbon atoms in 3 moles of $C_{2} H_{6} = 2 \times 3 = 6$ (ii) 1 mole of $C_{2} H_{6}$ contains 6 moles of hydrogen atoms. ∴ Number of moles of carbon atoms in 3 moles of $C_{2} H_{6} = 3 \times 6 = 18$ (iii) 1 mole of $C_{2} H_{6}$ contains $6.023 \times 1 0^{23}$ molecules of ethane. Number of molecules in 3 moles of $C_{2} H_{6}$ $= 3 \times 6.023 \times 1 0^{23} = 18.069 \times 1 0^{23}$ .

What is the concentration of sugar $(C_{12} H_{22} O_{11})$ in $mol L^{- 1}$ if its 20 g are dissolved in enough water to make a final volume up to 2 L ? Ans. Molarity (M) of a solution is given by,

= \frac{Numbers of moles of solute}{Volume of solution in Litres} = \frac{Mass of sugar / molar of sugar}{2 L}

= \frac{20 g / [( 12 \times 12 ) + ( 1 \times 22 ) + ( 11 \times 16 )] g}{2 L} = \frac{20 g /342 g}{2 L} = \frac{0.0585 mol}{2 L} = 0.02925 mol L^{- 1}

∴ Molar concentration of sugar $= 0.02925 mol L^{- 1}$

If the density of methanol is $0.793 kg L^{- 1}$ , what is its volume needed for making 2.5 L of its 0.25 M solution? Ans. Molar mass of methanol $(CH_{3} OH)$

= (1 \times 12) + (4 \times 1) + (1 \times 16) = 32 g mol^{- 1} = 0.032 kg mol^{- 1}

Molarity of methanol solution

= \frac{0.793 kg L ^{- 1}}{0.032 kg mol ^{- 1}} = 24.78 mol L^{- 1}

(Since density is mass per unit volume) Applying, $M_{1} V_{1} = M_{2} V_{2}$ (Given solution) (Solution to be prepared)

Pressure is determined as force per unit area of the surface. The SI unit of pressure, Pascal is as shown below: $1 Pa = 1 N m^{- 2}$ If mass of air at sea level is $1034 g cm^{- 2}$ , calculate the pressure in Pascal. Ans. Pressure is defined as force acting per unit area of the surface.

P = \frac{F}{A} = \frac{1034 g \times 9.8 ms ^{- 2}}{cm ^{2}} \times \frac{1 kg}{1000 g} \times \frac{( 100 ) ^{2} cm ^{2}}{1 m ^{2}} = 1.01332 \times 1 0^{5} kg m^{- 1} s^{- 2}

We know,

1 N = 1 kg ms^{- 2}

Then,

What is the SI unit of mass? How is it defined? Ans. The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of the international prototype of kilogram.
Match the following prefixes with their multiples:

Prefixes	Multiples
micro	10⁶
deca	10⁹
mega	10⁻⁶
giga	10⁻¹⁵
femto	10

Ans.

Prefixes	Multiples
micro	10⁶
deca	10
mega	10⁶
giga	10⁹
femto	10⁻¹⁵

What do you mean by significant figures? Ans. Significant figures are those meaningful digits that are known with certainty. They indicate uncertainty in an experiment or calculated value. For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures are 3. Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result.
A sample of drinking water was found to be severely contaminated with chloroform, $CHCl_{3}$ , supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in percent by mass. (ii) Determine the molality of chloroform in the water sample.

Ans.

(i) 1 ppm is equivalent to 1 part out of 1 million $(1 0^{6})$ parts. Mass percent of 15 ppm chloroform in water

= \frac{15}{1 0 ^{6}} \times 100 ≃ 1.5 \times 1 0^{- 3} %

(ii) 100 g of the sample contains $1.5 \times 1 0^{- 3} g$ of $CHCl_{3}$ . $\Rightarrow 1000 g$ of the sample contains $1.5 \times 1 0^{- 2} g$ of $CHCl_{3}$ . Molality of chloroform in water

= \frac{1.5 \times 1 0 ^{- 2} g}{Molar mass of CHCl _{3}}

Molar mass of $CHCl_{3}$

= 12.00 + 1.00 + 3 (35.5) = 119.5 g mol^{- 1}

Molality of chloroform in water

Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Ans. (i) $0.0048 = 4.8 \times 1 0^{- 3}$

(ii) $234, 000 = 2.34 \times 1 0^{5}$ (iii) $8008 = 8.008 \times 1 0^{3}$ (iv) $500.0 = 5.000 \times 1 0^{2}$ (v) $6.0012 = 6.0012$

How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034

Ans. (i) 0.0025 There are 2 significant figures.

(ii) 208 There are 3 significant figures. (iii) 5005 There are 4 significant figures. (iv) 126,000 There are 3 significant figures. (v) 500.0 There are 4 significant figures. (vi) 2.0034 There are 5 significant figures.

Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808 Ans. (i) 34.2 (ii) 10.4 (iii) 0.0460 (iv) 2810
The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

Mass of dinitrogen	Mass of dioxygen
14 g	16 g
14 g	32 g
28 g	32 g
28 g	80 g

(a) Which law of chemical combination is obeyed

by the above experimental data?

Give its statement.

(b) Fill in the blanks in the following conversions:

(i) 1 km = ...................... mm

= ...................... pm

(ii) 1 mg = ...................... kg

= ...................... ng

(iii) 1 mL = ...................... L

= ...................... dm3

Ans.

(a) If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are $32 g, 64 g, 32 g$ , and 80 g. The masses of dioxygen bear a whole number ratio of 1 : 2 : 2 : 5. Hence, the given experimental data obeys the law of multiple proportions. The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers. (b) (i) $1 km = 1 km \times = \frac{1000 m}{1 km} \times \frac{100 cm}{1 m} \times \frac{10 mm}{1 cm}$

∴ 1 km = 1 0^{6} mm 1 km = 1 km \times = \frac{1000 m}{1 km} \times \frac{1 pm}{1 0 ^{- 12}} 1 km = 1 0^{15} pm

Hence, $1 km = 1 0^{6} mm = 1 0^{15} pm$ (ii)

1 mL = 1 0^{- 3} dm^{3}

∴

If the speed of light is $3.0 \times 1 0^{8} m s^{- 1}$ , calculate the distance covered by light in 2.00 ns. Ans. According to the question: Time taken to cover the distance

= 2.00 ns = 2.00 \times 1 0^{- 9} s

Speed of light $= 3.0 \times 1 0^{8} ms^{- 1}$ Distance travelled by light in 2.00 ns

In a reaction

A + B_{2} \to AB_{2}

Identify the limiting reagent, if any, in the following reaction mixtures (i) 300 atoms of A + 200 molecules of B (ii) $2 mol A + 3 mol B$ (iii) 100 atoms of $A + 100$ molecules of B (iv) $5 mol A + 2.5 mol B$ (v) $2.5 mol A + 5 mol B$

Ans. A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.

(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent. (ii) According to the reaction, 1 mol of A reacts with 1 mol of B . Thus, 2 mol of A will react with only 2 mol of B. As a result, 1mol of B will not be consumed. Hence, A is the limiting reagent. (iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present. (iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent. (v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N_{2 (g)} + H_{2 (g)} \to 2 NH_{3 (g)}

(i) Calculate the mass of ammonia produced if $2.00 \times 1 0^{3} g$ dinitrogen reacts with $1.00 \times$ $1 0^{3} g$ of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?

Ans.

(i) Balancing the given chemical equation,

N_{2 (g)} + 3 H_{2 (g)} \to 2 NH_{3 (g)}

From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia. $\Rightarrow 2.00 \times 1 0^{3} g$ of dinitrogen will react with $\frac{6 g}{28 g} \times 2.00 \times 1 0^{3} g$ dihydrogen i.e., $2.00 \times 1 0^{3} g$ of dinitrogen will react with 428.6 g of dihydrogen. Given, Amount of dihydrogen $= 1.00 \times 1 0^{3} g$ Hence, $N_{2}$ is the limiting reagent. $∴ 28 g$ of $N_{2}$ produces 34 g of $NH_{3}$ . Hence, mass of ammonia produced by

2000 g of N_{2} = \frac{34 g}{28 g} \times 2000 g = 2428.57 g

(ii) $N_{2}$ is the limiting reagent and $H_{2}$ is the excess reagent. Hence, $H_{2}$ will remain unreacted. (iii) Mass of dihydrogen left unreacted

How are 0.50 mol $Na_{2} CO_{3}$ and 0.50 $MNa_{2} CO_{3}$ different? Ans. Molar mass of $Na_{2} CO_{3}$

= (2 \times 23) + 12.00 + (3 \times 16) = 106 g mol^{- 1}

Now, 1 mole of $Na_{2} CO_{3}$ means 106 g of $Na_{2} CO_{3}$ . $∴ 0.5 mol$ of $Na_{2} CO_{3}$

= \frac{106 g}{1 mole} \times 0.5 mol Na_{2} CO_{3} = 53 g Na_{2} CO_{3}

⇒

0.50 M of Na_{2} CO_{3} = 0.50 mol / L Na_{2} CO_{3}

Hence, 0.50 mol of $Na_{2} CO_{3}$ is present in 1 L of water or 53 g of $Na_{2} CO_{3}$ is present in 1 L of water.

If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced? Ans. Reaction of dihydrogen with dioxygen can be written as:

2 N_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(g)}

Now, two volumes of dihydrogen react with one volume of dioxygen to produce two volumes of water vapour. Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg

Ans.

(i) 28.7 pm :

1 pm = 1 0^{- 12} m ∴ 28.7 pm = 28.7 \overset{˚}{A} - 1 0^{- 12} m = 2.87 \overset{˚}{A} - 1 0^{- 11}

(ii) 15.15 pm :

1 pm = 1 0^{- 12} m ∴ 15.15 pm = 15.15 \overset{˚}{A} - 1 0^{- 12} m = 1.515 \overset{˚}{A} - 1 0^{- 11} m

(iii) 25365 mg :

1 mg = 1 0^{- 3} g 25365 mg = 2.5365 \overset{˚}{A} - 1 0^{4} \overset{˚}{A} - 1 0^{- 3} g

Since,

1 g = 1 0^{- 3} kg 2.5365 \overset{˚}{A} - 1 0^{1} g = 2.5365 \overset{˚}{A} - 1 0^{1} \overset{˚}{A} - 1 0^{- 3} kg

$∴ 25365 mg = 2.5365 \overset{˚}{A} - 1 0^{- 2} kg$

Which one of the following will have largest number of atoms? (i) 1 g Au (s) (ii) $1 g Na (s)$ (iii) 1 g Li (s) (iv) 1 g of $Cl_{2} (g)$ Ans. 1 g of $Au (s) = \frac{1}{197} mol$ of $Au (s)$

= \frac{6.022 \times 1 0 ^{23}}{197} atoms of Au (s) = 3.06 \overset{˚}{A} - 1 0^{21} atoms of Au (s)

1 g of Na (s) = \frac{1}{23} mol of Na (s) = \frac{6.022 \times 1 0 ^{23}}{23} atoms of Na (s) = 0.262 \overset{˚}{A} - 1 0^{23} atoms of Na (s) = 26.2 \overset{˚}{A} - 1 0^{21} atoms of Na (s)

1 g of Li (s) $= \frac{1}{7} mol$ of Li (s)

= \frac{6.022 \times 1 0 ^{23}}{7} atoms of Li (s) = 0.86 \overset{˚}{A} - 1 0^{23} atoms of Li (s) = 86.0 \overset{˚}{A} - 1 0^{21} atoms of Li (s)

1 g of $Cl_{2} (g) = \frac{1}{71} mol$ of $Cl_{2} (g)$ (Molar mass of $Cl_{2}$ molecule

= 35.5 \overset{˚}{A} - 2 = 71 g mol^{- 1}) = \frac{6.022 \times 1 0 ^{23}}{71} molecules of Cl_{2} (g) = 0.0848 \overset{˚}{A} - 1 0^{23} molecules of Cl_{2} (g) = 8.48 \overset{˚}{A} - 1 0^{21} molecules of Cl_{2} (g)

As one molecule of $Cl_{2}$ contains two atoms of Cl. Number of atoms of $Cl = 2 \overset{˚}{A} - 8.48 \overset{˚}{A} - 1 0^{21}$

= 16.96 \overset{˚}{A} - 1 0^{21} atoms of Cl

Hence, 1 g of $Li (s)$ will have the largest number of atoms.

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). Ans. Mole fraction of $C_{2} H_{5} OH$

= \frac{Number of moles of C _{2} H _{5} OH}{Number of moles of solution} 0.040 = \frac{n _{C_{2} H_{5} OH}}{n _{C_{2} H_{5} OH} + n _{H_{2} O}} \dots\dots\dots\dots\dots ..

Number of moles present in 1 L water :

n_{H_{2} O} n_{H_{2} O} = \frac{1000 g}{18 g mol ^{- 1}} = 55.55 mol

Substituting the value of $n_{H_{2} O}$ in equation (1),

What will be the mass of one $^{12} C$ atom in g ? Ans. 1 mole of carbon atoms $= 6.023 \times 1 0^{23}$ atoms of carbon $= 12 g$ of carbon
How many significant figures should be present in the answer of the following calculations?

Ans.

(i) $\frac{0.02856 \times 298.15 \times 0.112}{0.5785}$ (ii) 5 × 5.364 (iii) $0.0125 + 0.7864 + 0.0215$

Ans.

(i) $\frac{0.02856 \times 298.15 \times 0.112}{0.5785}$ Least precise number of calculation $= 0.112$ ∴ Number of significant figures in the answer = Number of significant figures in the least precise number = 3 (ii) 5 × 5.364 Least precise number of calculation = 5.364 ∴ Number of significant figures in the answer = Number of significant figures in $5.364 = 4$ (iii) $0.0125 + 0.7864 + 0.0215$ Since the least number of decimal places in each term is four, the number of significant figures in the answer is also 4.

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope	Isotope molar mass	Abundance
³⁶Ar	35.96755 g mol⁻¹	0.337 %
³⁸Ar	37.96272 g mol⁻¹	0.063 %
⁴⁰Ar	39.9624 g mol⁻¹	99.600 %

Ans. Molar mass of argon

Calculate the number of atoms in each of the following : (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

Ans.

(i) 1 mole of $Ar = 6.022 \times 1 0^{23}$ atoms of Ar $∴ 52 mol$ of $Ar = 52 \times 6.022 \times 1 0^{23}$ atoms of Ar $= 3.131 \times 1 0^{25}$ atoms of Ar (ii) 1 atom of He = 4 u of He Or, 4 u of $He = 1$ atom of He 1 u of He $\frac{1}{4}$ atom of He 52 u of $He = \frac{52}{4}$ atom of He $= 13$ atoms of He (iii) 4 g of $He = 6.022 \times 1 0^{23}$ atoms of He $∴ 52 g$ of $He = \frac{6.022 \times 1 0 ^{23} \times 52}{4}$ atoms of $He = 7.8286 \times 1 0^{24}$ atoms of He

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g .

Calculate :

(i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Ans.

(i) 1 mole (44 g) of $CO_{2}$ contains 12 g of carbon. $∴ 3.38 g$ of $CO_{2}$ will contain carbon

= \frac{12 g}{44 g} \times 3.38 g = 0.9217 g

18 g of water contains 2 g of hydrogen. $∴ 0.690 g$ of water will contain hydrogen

= \frac{2 g}{18 g} \times 0.690 g = 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g = 0.9984 g

$∴$ Percent of C in the compound

= \frac{0.9217 g}{0.9984 g} \times 100 = 92.32%

Percent of H in the compound

= \frac{0.0767 g}{0.9984 g} \times 100 = 7.68%

Moles of carbon in the compound

= \frac{92.32}{12.00} = 7.69

Moles of hydrogen in the compound

= \frac{7.68}{1} = 7.68

∴ Ratio of carbon to hydrogen in the compound $= 7.69 : 7.68 = 1 : 1$ Hence, the empirical formula of the gas is CH. (ii) Given, Weight of 10.0 L of the gas (at S.T.P) $= 11.6 g$ ∴ Weight of 22.4 L of gas at STP

= \frac{11.6 g}{10.0 L} \times 22.4 L = 25.984 g ≃ 26 g

Hence, the molar mass of the gas is 26 g. (iii) Empirical formula mass of $CH = 12 + 1 = 13 g$ $n = \frac{Molar mass of gas}{Empirical formula mass of gas} = \frac{26 g}{13 g}$ $n = 2$ ∴ Molecular formula of gas $= (CH)_{n} = C_{2} H_{2}$

Calcium carbonate reacts with aqueous HCl to give $CaCl_{2}$ and $CO_{2}$ according to the reaction, $CaCO_{3} (s) + 2 HCl_{(aq)} \to CaCl_{2 (aq)} + CO_{2 (g)}$ $+ H_{2} O_{(l)}$ What mass of $CaCO_{3}$ is required to react completely with 25 mL of 0.75 M HCl ? Ans. 0.75 M of $HCl \equiv 0.75 mol$ of HCl are present in 1 L of water $\equiv [(0.75 mol) \times (36.5 g mol)^{- 1})] HCl$ is present in 1 L of water $\equiv 27.375 g$ of HCl is present in 1 L of water Thus, 1000 mL of solution contains 27.375 g of HCl . ∴ Amount of HCl present in 25 mL of solution

= \frac{27.375 g}{1000 mL} \times 25 mL = 0.6844 g

From the given chemical equation, 2 mol of $HCl (2 \times 36.5 = 73 g)$ react with 1 mol of $CaCO_{3} (100 g)$ . Amount of $CaCO_{3}$ that will react with 0.6844

Chlorine is prepared in the laboratory by treating manganese dioxide ( $MnO_{2}$ ) with aqueous hydrochloric acid according to the reaction

4 HCl_{(aq)} + MnO_{2 (s)} \to 2 H_{2} O_{(l)} + MnCl_{2 (aq)} + Cl_{2 (g)}

How many grams of HCl react with 5.0 g of manganese dioxide? Ans. $1 mol [55 + 2 \times 16 = 87 g] MnO_{2}$ reacts completely with $4 mol [4 \times 36.5 = 146 g]$ of HCl. 5.0 g of $MnO_{2}$ will react with

= \frac{146 g}{87 mL} \times 5.0 g of HCl = 8.4 g of HCl

Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

Explore NCERT Solutions for Class 11 Chemistry with clear answers to textbook questions from every chapter. Understand important concepts, revise key topics, and strengthen your Chemistry basics for school exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 2	Structure of Atom
Chapter 3	Classification of Elements and Periodicity in Properties
Chapter 4	Chemical Bonding and Molecular Structure
Chapter 5	Chemical Thermodynamics
Chapter 6	Equilibrium
Chapter 7	Redox Reactions
Chapter 8	Organic Chemistry – Some Basic Principles and Techniques
Chapter 9	Hydrocarbons

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 1

Step-by-Step Numerical Solutions: Each problem is explained in clear steps, with unit conversions shown to help students avoid calculation errors.
Accurate Calculations: Solutions explain significant figures, scientific notation, and rounding off to help students write accurate answers.
Stoichiometry and Limiting Reagent: Simple methods help students solve numerical problems based on chemical equations, limiting reagents, and the quantities of reactants and products.
Mole Concept and Gas Volume: Clear explanations cover the mole concept and the relationship between the amount of a gas, its volume, temperature, and pressure. At STP, the molar volume of an ideal gas is approximately 22.7 L when STP is taken as 1 bar and 273.15 K.
Solutions by ALLEN Faculty: NCERT Solutions prepared by ALLEN faculty help students understand the chapter's concepts and practise questions based on the Class 11 Chemistry syllabus.

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NCERT Solutions Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry

1.0Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 1

NCERT EXERCISE

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 1

Frequently Asked Questions

What topics are covered in NCERT Solutions for Class 11 Chemistry Chapter 1?

How do NCERT Solutions for Class 11 Chemistry Chapter 1 explain the laws of chemical combination?

Why is the mole concept important in NCERT Solutions for Class 11 Chemistry Chapter 1?

How do NCERT Solutions for Class 11 Chemistry Chapter 1 explain empirical and molecular formulas?

Are NCERT Solutions for Class 11 Chemistry Chapter 1 important for board and competitive exams?

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Frequently Asked Questions

What topics are covered in NCERT Solutions for Class 11 Chemistry Chapter 1?

How do NCERT Solutions for Class 11 Chemistry Chapter 1 explain the laws of chemical combination?

Why is the mole concept important in NCERT Solutions for Class 11 Chemistry Chapter 1?

How do NCERT Solutions for Class 11 Chemistry Chapter 1 explain empirical and molecular formulas?

Are NCERT Solutions for Class 11 Chemistry Chapter 1 important for board and competitive exams?

Join ALLEN!

NCERT Solutions Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry

1.0Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 1NCERT EXERCISE

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 1

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 1

NCERT EXERCISE