Class 11 Chemistry Chapter 1, Some Basic Concepts of Chemistry, introduces the fundamental ideas students need to understand before studying advanced chemistry. The chapter explains how matter is measured, how atoms and molecules are counted, and how the quantities of reactants and products are calculated in a chemical reaction.
Students learn important concepts such as atomic mass, molecular mass, mole concept, molar mass, percentage composition, empirical formula, molecular formula, concentration terms, stoichiometry, significant figures and limiting reagents. These concepts are used repeatedly in physical, inorganic and organic chemistry.
The NCERT solutions provided below explain the main numerical problems in a simple, step-by-step format. Students can use them to complete homework, revise the chapter and prepare for CBSE school examinations, JEE Main and NEET.
Class 11 Chemistry Chapter 1 Overview
Some Basic Concepts of Chemistry establishes a relationship between the microscopic particles present in matter and the measurable quantities used in a laboratory.
The chapter begins with the importance of chemistry and the nature of matter. It then introduces SI units, scientific notation, significant figures and uncertainty in measurement. Students also study the laws of chemical combination and Dalton’s atomic theory.
The second half of the chapter focuses on calculations involving atomic masses, molecular masses, moles, percentage composition, empirical and molecular formulae, chemical equations and limiting reagents. These topics are included in the official Class 11 Chemistry syllabus.
NCERT Solutions For Class 11 Chemistry chapter 1-Some Basic Concepts of Chemistry
Answer the following Questions.
- Calculate the molecular mass of the following :
(i) H₂O
(ii) CO₂
(iii) CH₄
Solution :
(i)CH₄ :
Molecular weight of methane, CH₄
= (1 × Atomic weight of carbon) + (4 × Atomic weight of hydrogen)
= [1(12.011 u) +4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u
(ii) H₂O :
Molecular weight of water, H₂O
= (2 × Atomic weight of hydrogen) + (1 × Atomic weight of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u +16.00 u
= 18.016 u
So approximately
= 18.02 u
(iii) CO₂ :
= Molecular weight of carbon dioxide, CO₂
= (1 × Atomic weight of carbon) + (2 × Atomic weight of oxygen)
= [1(12.011 u) + 2(16.00 u)]
= 12.011 u +32.00 u
= 44.011 u
So approximately
= 44.01 u
- Calculate the mass percent of different elements present in Sodium Sulphate (Na₂SO₄).
Solution :
Molar mass of Na₂SO₄ = [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 g
Mass percent of an element = (Mass of that element in compound/Molar mass of that compound) × 100
∴ Mass percent of sodium (Na): (46/142) × 100 = 32.39%
Mass percent of sulphur (S): (32/142) × 100 = 22.54%
Mass percent of oxygen (O): (64/142) × 100 = 45.07%
- Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Solution :
% of iron by mass = 69.9% [Given]
% of oxygen by mass = 30.1% [Given]
Atomic mass of iron = 55.85 amu
Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = % mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = % mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25
⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula of the iron oxide is Fe₂O₃.
- 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.
Solution :
The balanced reaction of combustion of carbon in dioxygen is:
C(s) + O₂(g) → CO₂(g)
1 mole 1 mole (32 g) 1 mole (44 g)
(i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide produced by burning 1 mole of carbon.
(ii) Here, oxygen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide.
(iii) Here again oxygen acts as a limiting reagent as 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 (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. The molar mass of sodium acetate is 82.0245 g mol⁻¹.
Solution :
0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles of sodium acetate.
∴ Number of moles of sodium acetate in 500 mL = (0.375/1000) × 500 = 0.375/2 = 0.1875
Molar mass of sodium acetate = 82.0245 g mol⁻¹
∴ Mass of sodium acetate required = 0.1875 × 82.0245 g = 15.380 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%.
Solution :
Mass percent of 69% means that 100 g of nitric acid solution contains 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO₃) = 1 + 14 + 48 = 63 g mol⁻¹
Number of moles in 69 g of HNO₃ = 69/63 moles = 1.095 moles
Volume of 100 g nitric acid solution = 100/1.41 mL = 70.92 mL = 0.07092 L
∴ Concentration of HNO₃ in moles per litre = 1.095/0.07092 = 15.44 M
- How much copper can be obtained from 100 g of copper sulphate (CuSO₄ )?
Solution : 1 mole of CuSO₄ contains 1 mole of copper.
Molar mass of CuSO₄ = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 g
159.5 g of CuSO₄ contains 63.5 g of copper.
∴ Copper can be obtained from 100 g of copper sulphate = (63.5/159.5) × 100 = 39.81 g
- 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
Solution :
% of iron by mass = 69.9% [Given]
% of oxygen by mass = 30.1% [Given]
Atomic mass of iron = 55.85 amu
Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = % mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = % mass of oxygen by mass /Atomic mass of oxygen = 30.01/16 =1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25
⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula of the iron oxide is Fe₂O₃.
Mass of Fe₂O₃ = (2 × 55.85) + (3 × 16.00) = 159.7 g mol⁻¹
n = Molar mass /Empirical formula mass = 159.7/ 159.6 = 1 (approx.)
Thus, Molecular formula is same as empirical formula i.e. Fe₂O₃
- Calculate the atomic mass (average) of chlorine using the following data :
| % Natural Abundance | Molar Mass | |
| 35Cl | 75.77 | 34.9689 |
| 37Cl | 24.23 | 36.9659 |
Solution :
Fractional Abundance of ³⁵Cl = 0.7577 and Molar mass = 34.9689
Fractional Abundance of ³⁷Cl = 0.2423 and Molar mass = 36.9659
∴ Average atomic mass = (0.7577 × 34.9689) amu + (0.2423 × 36.9659)
= 26.4959 + 8.9568 = 35.4527
- In three moles of ethane (C₂H₆), calculate the following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution :
(i) 1 mole of C₂H₆ contains 2 moles of Carbon atoms
∴ 3 moles of C₂H₆ will contain 6 moles of Carbon atoms
(ii) 1 mole of C₂H₆ contains 6 moles of Hydrogen atoms
∴ 3 moles of C₂H₆ will contain 18 moles of Hydrogen atoms
(iii) 1 mole of C₂H₆ contains Avogadro’s no. 6.022 × 10²³ molecules
∴ 3 moles of C₂H₆ will contain ethane molecule = 3 × 6.022 × 10²³ = 1.8066 × 10²⁴ molecules.
- What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L⁻¹ if its 20 g are dissolved in enough water to make a final volume up to 2 L?
Solution :
Molar mass of sugar (C₁₂H₂₂O₁₁) = (12 ×12) +(1 ×22)+ (11 × 16) = 342 g mol⁻¹
Number of moles in 20 g of sugar = 20/342 = 0.0585 mole
Volume of Solution = 2 L (given)
Molar concentration = Moles of solute/Volume of solution in L = 0.0585 mol /2 L = 0.0293 mol L⁻¹ = 0.0293 M
- 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?
Solution :
Molar mass of methanol (CH₃OH) = (1 × 12) + (4 × 1) + (1 × 16) = 32 g mol⁻¹ = 0.032 kg mol⁻¹
Molarity of the solution = 0.793 /0.032 = 24.78 mol L⁻¹
Applying, M₁V₁ (Given Solution) = M₂V₂ (Solution to be prepared)
24.78 × V₁ = 0.25 × 2.5 L
V₁= 0.02522 L = 25.22 mL
- 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 the mass of air at sea level is 1034 g cm⁻²,calculate the pressure in pascal.
Solution :
Pressure is the force (i.e. weight) acting per unit area.
P = F/A = (1034 g cm⁻²) × 9.8 m s⁻²
= 1034 g cm⁻² × 9.8 m s⁻² × (1 kg/1000 g) × (10⁴ cm²/1 m²)
= 1.01332 × 10⁵ N m⁻²
Now,
1 Pa = 1 N m⁻²
∴ 1.01332 × 10⁵ N m⁻² = 1.01332 × 10⁵ Pa
- What is the SI unit of mass? How is it defined?
Solution : The SI unit of mass is kilogram (kg).
The kg is defined as the mass of a platinum-iridium (Pt-Ir) cylinder that is stored in an air-tight jar at the International Bureau of Weigh and Measures in France.
- Match the following prefixes with their multiples:
Prefixes
Multiples
(a) femto 10⁻¹⁵
(b) giga 10⁹
(c) mega 10⁶
(d) deca 10¹
(e) micro 10⁻⁶
Solution :
Prefixes Multiples
(a) femto — 10⁻¹⁵
(b) giga — 10⁹
(c) mega — 10⁶
(d) deca — 10¹
(e) micro — 10⁻⁶
- What do you mean by significant figures ?
Solution :
Significant figures are meaningful digits which are known with certainty including the last digit whose value is uncertain.
For example,
In 11.2546 g, there are 6 significant figures but here 11.254 is certain and 6 is uncertain and the uncertainty would be ±1 in the last digit. Hence the last uncertain digit is also included in Significant figures.
- A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, 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.
Solution :
(i) 15 ppm means 15 parts in 10⁶ parts.
∴ % by mass = 15/10⁶ × 100 = 15 × 10⁻⁴ = 1.5 × 10⁻³%
(ii) Molar mass of chloroform (CHCl₃) = 12 + 1+ (3 × 35.5) = 118.5 g mol⁻¹
100 g of the sample contain chloroform = 1.5 × 10⁻³ g
∴ 1000 g (1 kg) of the sample will contain chloroform = 1.5 × 10⁻² g
= 1.5 × 10⁻²/ 118.65 mole = 1.266 × 10⁻⁴ mole
∴ Molality = 1.266 × 10⁻⁴ m.
- Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Solution :
(i) 0.0048 = 4.8 × 10⁻³
(ii) 234, 000 = 2.34 × 10⁵
(iii) 8008 = 8.008 × 10³
(iv) 500.0 = 5.000 × 10²
(v) 6.0012 = 6.0012 × 10⁰
- 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
Solution :
(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5
- Round up the following up to three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Solution :
(i) 34.2
(ii) 10.4
(iii) 0.046
(iv) 2810
- The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dinitrogen Mass of dioxygen
(i)14 g 16 g
(ii)14 g 32 g
(iii)28 g 32 g
(iv)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
Solution :
(a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides. These masses of dioxygen bear a simple whole number ratio as 2:4:2:5. Hence, the data given will obey the law of multiple proportions.
The statement is as follows: two elements always combined in a fixed mass of other bearing a simple ratio to another to form two or more chemical compounds.
(b) (i) 1 km = 1 km × (10³ m/1 km) × (10³ mm/1 m) = 10⁶ mm
1 km = 1 km × (10³ m/1 km) × (1 pm/10⁻¹² m) = 10¹⁵ pm
(ii) 1 mg = 1 mg × (1 g/10³ mg) × (1 kg/10³ g) = 10⁻⁶ kg
1 mg = 1 mg × (1 g/10³ mg) × (1 ng/10⁻⁹ g) = 10⁶ ng
(iii) 1 mL = 1 mL × (1 L/10³ mL) = 10⁻³ L
1 mL = 1 cm³ = 1 cm³ × (1 dm/10 cm)³ = 10⁻³ dm³
- If the speed of light is 3.0 × 10⁸ m s⁻¹, calculate the distance covered by light in 2.00 ns.
Solution :
Distance covered = Speed × Time = 3.0 × 10⁸ m s⁻¹ × 2.00 ns
= 3.0 × 10⁸ m s⁻¹ × 2.00 ns × 10⁻⁹ s /1 ns = 6.00 × 10⁻¹ m = 0.600 m
- In a reaction
A + B₂ → AB₂
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B₂
(ii) 2 mol A + 3 mol B₂
(iii) 100 atoms of A + 100 molecules of B₂
(iv) 5 mol A + 2.5 mol B₂
(v) 2.5 mol A + 5 mol B₂
Solution :
(i) According to the reaction, 1 atom of A reacts with 1 molecule of B₂.
∴ 200 molecules of B₂ will react with 200 atoms of A, thereby leaving 100 atoms of A unreacted. Hence, B₂ is the limiting reagent.
(ii) According to the reaction, 1 mol of A reacts with 1 mol of B₂.
∴ 2 mol of A will react with only 2 mol of B₂ leaving 1 mol of B₂. Hence, A is the limiting reagent.
(iii) 1 atom of A combines with 1 molecule of B₂.
∴ All 100 atoms of A will combine with all 100 molecules of B₂. Hence, the mixture is stoichiometric and there is no limiting reagent.
(iv) 1 mol of atom A combines with 1 mol of molecule B₂.
∴ 2.5 mol of B₂ will combine with only 2.5 mol of A. and 2.5 mol of A will be left unreacted. Hence, B₂ is the limiting reagent.
(v) 1 mol of atom A combines with 1 mol of molecule B₂.
∴ 2.5 mol of A will combine with only 2.5 mol of B₂ and the remaining 2.5 mol of B₂ will be left. Hence, A is the limiting reagent.
- Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N₂(g) + 3H₂(g) → 2NH₃(g)
(i) Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 × 10³ g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Solution :
1 mole of dinitrogen (28 g) reacts with 3 moles of dihydrogen (6 g) to give 2 moles of ammonia (34 g).
∴ 2000 g of N₂ will react with H₂ = 6/28 × 2000 g = 428.6 g. Thus, here N₂ is the limiting reagent while H₂ is in excess.
28 g of N₂ produce 34 g of NH₃.
∴ 2000 g of N₂ will produce = 34/28 × 2000 g = 2428.57 g of NH₃.
(ii) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ will remain unreacted.
(iii) Mass of dihydrogen left unreacted = 1000 g – 428.6 g = 571.4
- How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Solution :
Molar mass of Na₂CO₃ = (2 × 23) +12.00+(3 × 16) = 106 g mol⁻¹
∴ 0.50 mol Na₂CO₃ means 0.50 × 106 g = 53 g
0.50 M Na₂CO₃ means 0.50 mol of Na₂CO₃ i.e. 53 g of Na₂CO₃ are present in 1 litre of the solution.
- If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Solution :
Dihydrogen gas reacts with dioxygen gas as,
2H₂(g) + O₂(g) → 2H₂O(g)
Thus, 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
Solution :
(i) 1 pm = 10⁻¹² m
28.7 pm = 28.7 × 10⁻¹² m = 2.87 × 10⁻¹¹ m
(ii) 1 pm = 10⁻¹² m
∴ 15.15 pm = 15.15 × 10⁻¹² m = 1.515 × 10⁻¹¹ m
(iii) 1 mg = 10⁻³ g
25365 mg = 25.365 g
Now,
1 g = 10⁻³ kg
25.365 g = 25.365 × 10⁻³ kg
∴ 25365 mg = 2.5365 × 10⁻² 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 Cl₂(g)
Solution :
(i) 1 g Au = 1/197 mol = 1/197 × 6.022 × 10²³ atoms
(ii) 1 g Na = 1/23 mol = 1/23 × 6.022 × 10²³ atoms
(iii) 1 g Li = 1/7 mol = 1/7 × 6.022 × 10²³ atoms
(iv) 1 g Cl₂ = 1/71 mol = 1/71 × 6.022 × 10²³ atoms
Thus, 1 g of Li has 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).
Solution :
Mole fraction of C₂H₅OH = Number of moles of C₂H₅OH /Number of moles of solution
x(C₂H₅OH) = n(C₂H₅OH)/[n(C₂H₅OH) + n(H₂O)] = 0.040 …(1)
We have to find the number of moles of ethanol in 1 L of the solution but the solution is dilute. Therefore, water is approx. 1 L.
Number of moles in 1 L of water = 1000 g /18 g mol⁻¹ = 55.55 moles
Substituting n(H₂O) = 55.55 in equation 1
n(C₂H₅OH)/[n(C₂H₅OH) + 55.55] = 0.040
⇒ 0.96n (C₂H₅OH) = 55.55 × 0.040
⇒ n(C₂H₅OH) = 2.31 mol
Hence, molarity of the solution = 2.31 M
- What will be the mass of one ¹²C atom in g ?
Solution :
1 mol of ¹²C atoms = 6.022 × 10²³ atoms = 12 g
∴ Mass of 1 atom ¹²C = 12/(6.022 × 10²³) g = 1.9927 × 10⁻²³ g
- How many significant figures should be present in the answer of the following calculations?
(i) 0.02856 × 298.15 × 0.112 /0.5785
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Solution :
(i) Least precise term i.e. 0.112 is having 3 significant digits.
∴ There will be 3 significant figures in the calculation.
(ii) 5.364 is having 4 significant figures.
∴ There will be 4 significant figures in the calculation.
(iii) Least number of decimal places in each term is 4.
∴ There will be 4 significant figures in the calculation.
- Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
Isotope Isotopic 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%
Solution :
Molar mass of Ar = ∑piAi
= (0.00337 × 35.96755 )+ (0.00063 × 37.96272 )+(0.99600 × 39.9624 ) = 39.948 g mol⁻¹
- 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.
Solution :
(i) 1 mol of Ar = 6.022 × 10²³ atoms
∴ 52 mol of Ar = 52 × 6.022 × 10²³ atoms = 3.131 × 10²⁵ atoms
(ii) 1 atom of He = 4 u of He
4 u of He = 1 Atom of He
∴ 52 u of He = 1/4 × 52 = 13 atoms
(iii) 1 mol of He = 4 g = 6.022 × 10²³ atoms
∴ 52 g of He = (6.022 × 10²³/4) × 52 atoms = 7.8286 × 10²⁴ atoms
- 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.
Solution :
Amount of carbon in 3.38 g of CO₂ = 12/44 × 3.38 g = 0.9218 g
Amount of hydrogen in 0.690 g H₂O = 2/18 × 0.690 g = 0.0767 g
The compound contains only C and H, therefore total mass of the compound = 0.9218 + 0.0767 = 0.9985 g
% of C in the compound = (0.9218 /0.9985 ) × 100 = 92.32
% of H in the compound = (0.0767 /0.9985 ) × 100 = 7.68
(i) Calculation of empirical formula,
Moles of carbon in the compound = 92.32/12 = 7.69
Moles of hydrogen in the compound = 7.68/1 = 7.68
Simplest molar ratio = 7.69 : 7.68 = 1 (approx.)
∴ Empirical formula CH
(ii) 10.0 L of the gas at STP weigh = 11.6 g
∴ 22.4 L of the gas at STP = 11.6/10.0 × 22.4 = 25.984 = 26 (approx.)
∴ Molar mass of gas = 26 g mol⁻¹
(iii) Mass of empirical formula CH = 12 + 1 = 13
∴ n = Molecular Mass/empirical formula = 26/13 = 2
∴ Molecular formula = C₂H₂
- Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?
Solution :
1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75 × 36.5 g = 24.375 g
∴ Mass of HCl in 25 mL of 0.75 M HCl = 27.375/1000 × 25 g = 0.6844 g
From the given chemical equation,
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO₃ i.e. 100 g
∴ 0.6844 g HCl reacts completely with CaCO₃ = 100/73 × 0.6844 g = 0.938 g
- Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
4HCl(aq) + MnO₂(s) → 2H₂O(l) + MnCl₂(aq) + Cl₂(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Solution :
1 mol of MnO₂ = 55 + 32 g = 87 g
87 g of MnO₂ react with 4 moles of HCl i.e. 4 × 36.5 g = 146 g of HCl.
∴ 5.0 g of MnO₂ will react with HCl = 146/87 × 5.0 g = 8.40 g.
Why Class 11 Chemistry Chapter 1 Matters in NEET and JEE
Class 11 Chemistry Chapter 1 is important because it builds the foundation for almost every numerical and conceptual topic studied later in chemistry. Concepts such as the mole, molar mass, concentration, stoichiometry, significant figures and limiting reagents are regularly used in thermodynamics, equilibrium, redox reactions, solutions and electrochemistry. The chapter helps students understand how microscopic particles such as atoms and molecules are connected with measurable quantities like mass, volume and concentration. It also develops essential calculation, unit-conversion and problem-solving skills. Questions from the mole concept and stoichiometry are frequently asked in CBSE examinations, JEE and NEET. A strong understanding of this chapter therefore makes advanced chemistry topics easier to understand and improves accuracy in numerical problems.
Preparation Tips for Class 11 Chemistry Chapter 1
Begin by learning the basic formulae for moles, molar mass, molarity, molality and mole fraction. Always write the balanced chemical equation before attempting a stoichiometry problem.
Keep track of units at every step. Convert millilitres into litres and grams into kilograms whenever required. While solving empirical-formula questions, divide each percentage by the corresponding atomic mass and reduce the results to the simplest whole-number ratio.
Practise limiting-reagent questions by comparing the available mole ratio with the ratio given in the balanced equation. Finally, revise the rules of significant figures because incorrect rounding can lead to lost marks even when the calculation method is correct.
Frequently Asked Questions
What are the most important topics in Class 11 Chemistry Chapter 1?
The most important topics are the mole concept, molar mass, laws of chemical combination, significant figures, percentage composition, empirical formula, molecular formula, concentration terms, stoichiometry and limiting reagents.
Are NCERT solutions enough for the Class 11 Chemistry examination?
NCERT solutions are the primary resource for understanding the chapter and preparing for school examinations. Students should first complete all NCERT examples and exercises and then practise additional numerical problems to improve speed and confidence.
Why do students find the mole concept difficult?
The mole concept connects several quantities, including mass, number of particles, gas volume and concentration. Students often become confused when selecting the correct conversion. Writing the given data, required quantity and conversion formula separately makes such questions easier.
How can I identify the limiting reagent?
First, convert the quantities of all reactants into moles. Then compare their mole ratio with the coefficients in the balanced equation. The reactant that produces the smaller amount of product is the limiting reagent.
What is the difference between molarity and molality?
Molarity is the number of moles of solute present in one litre of solution. Molality is the number of moles of solute present in one kilogram of solvent. Molarity can change with temperature because solution volume changes, whereas molality is generally independent of temperature.
Is Some Basic Concepts of Chemistry important for JEE and NEET?
Yes. The chapter provides the calculation methods required in many physical, inorganic and organic chemistry problems. A strong understanding of mole concept and stoichiometry helps students solve both direct questions and multi-concept numerical problems in JEE and NEET.