CLASS 11 · CHEMISTRY · CHAPTER 1

Some Basic Concepts of Chemistry

Simple worked solutions to all 36 end-of-chapter exercises on measurement, mole calculations, chemical formulae, and stoichiometry.

36 worked answersComplete chapter answers

Read the NCERT questions ↗← All chapters

36 worked answers · short steps first, explanations when you need them

Chapter 1 Exercises

Textbook pages 25–28

Question 1

Calculate the molar mass of (i) H₂O (ii) CO₂ (iii) CH₄.

Steps only

Add the atomic masses of every atom in each formula.

Answer

(i) H₂O: 2(1.008)+16.00 = 18.02 g mol⁻¹.
(ii) CO₂: 12.01+2(16.00) = 44.01 g mol⁻¹.
(iii) CH₄: 12.01+4(1.008) = 16.04 g mol⁻¹.

Question 2

Find the mass percentage of each element in Na₂SO₄.

Steps only

Find the total molar mass: 142 g mol⁻¹.

Answer

Molar mass = 2(23)+32+4(16)=142 g mol⁻¹. Na: 46/142×100=32.4%; S: 32/142×100=22.5%; O: 64/142×100=45.1%.

Question 3

An iron oxide has 69.9% iron and 30.1% oxygen by mass. Find its empirical formula.

Steps only

Assume 100 g of oxide so the percentages become grams.

Answer

For 100 g: Fe = 69.9/55.85 = 1.25 mol; O = 30.1/16 = 1.88 mol. Divide by 1.25: Fe:O ≈ 1:1.5. Multiply by 2: Fe₂O₃.

Question 4

Find CO₂ produced when (i) 1 mol C burns in air (ii) 1 mol C burns in 16 g O₂ (iii) 2 mol C burns in 16 g O₂.

Steps only

Use C + O₂ → CO₂; the mole ratio is 1:1:1.

Answer

C+O₂→CO₂.

(i) O₂ is abundant: 1 mol CO₂.
(ii) 16 g O₂=0.5 mol, so 0.5 mol CO₂.
(iii) O₂ remains limiting, so again 0.5 mol CO₂.

Question 5

Find the mass of CH₃COONa needed for 500 mL of 0.375 M solution. Molar mass = 82.0245 g mol⁻¹.

Steps only

Convert 500 mL to 0.500 L.

Answer

Moles = 0.375×0.500 = 0.1875 mol. Mass = 0.1875×82.0245 = 15.38 g.

Question 6

Find the molarity of nitric acid if solution density is 1.41 g mL⁻¹ and acid is 69% by mass.

Steps only

Use 1 L of solution; its mass is 1000×1.41 = 1410 g.

Answer

Take 1 L solution: mass = 1410 g. HNO₃ mass = 0.69×1410 = 972.9 g. Moles = 972.9/63.0 = 15.4 mol. Molarity = 15.4 mol L⁻¹.

Question 7

How much copper can be obtained from 100 g CuSO₄?

Steps only

Each mole of CuSO₄ contains one mole of Cu.

Answer

M(CuSO₄)=63.5+32+64=159.5 g mol⁻¹. Copper fraction = 63.5/159.5. Copper mass = 100×63.5/159.5 ≈ 39.8 g.

Question 8

Find the molecular formula of an iron oxide containing 69.9% iron and 30.1% oxygen.

Steps only

Convert the percentages to moles as in Question 1.3.

Answer

The simplest mole ratio is Fe:O = 1:1.5, or 2:3, so the empirical formula is Fe₂O₃. This is the molecular formula unit for this ionic oxide.

Question 9

Calculate chlorine’s average atomic mass: ³⁵Cl 75.77% (34.9689), ³⁷Cl 24.23% (36.9659).

Steps only

Change each percentage to a decimal fraction.

Answer

Average = 0.7577(34.9689)+0.2423(36.9659)=35.45 u (approximately).

Question 10

In 3 mol ethane (C₂H₆), find (i) moles of carbon atoms (ii) moles of hydrogen atoms (iii) number of ethane molecules.

Steps only

Each C₂H₆ molecule has 2 C atoms and 6 H atoms.

Answer

(i) 3×2=6 mol C atoms.
(ii) 3×6=18 mol H atoms.
(iii) 3×6.022×10²³ = 1.81×10²⁴ molecules.

Question 11

Find the molarity when 20 g sucrose (C₁₂H₂₂O₁₁) is dissolved to make 2 L solution.

Steps only

Calculate sucrose molar mass.

Answer

Molar mass = 342 g mol⁻¹. Moles = 20/342 = 0.0585 mol. M = 0.0585/2 = 0.0292 mol L⁻¹.

Question 12

Methanol density is 0.793 kg L⁻¹. What volume is needed for 2.5 L of 0.25 M solution?

Steps only

Calculate required moles and mass of methanol.

Answer

Required moles = 0.25×2.5 = 0.625 mol. Mass = 0.625×32.0 = 20.0 g. Volume = 20.0/0.793 = 25.2 mL.

Question 13

Air has mass 1034 g cm⁻² at sea level. Find the pressure in Pa.

Steps only

Convert g cm⁻² to kg m⁻² (multiply by 10).

Answer

Mass per area = 1034 g cm⁻² = 10,340 kg m⁻². Pressure = (mass/area)g = 10,340×9.8 ≈ 1.01×10⁵ Pa.

Question 14

What is the SI unit of mass and how is it defined?

Steps only

Use the modern SI definition.

Answer

The SI unit is the kilogram (kg). It is defined by fixing the numerical value of the Planck constant, h, at 6.62607015×10⁻³⁴ J s.

Question 15

Match micro, deca, mega, giga, and femto with powers of ten.

Steps only

Recall each standard SI prefix.

Answer

micro = 10⁻⁶; deca = 10¹; mega = 10⁶; giga = 10⁹; femto = 10⁻¹⁵.

Question 16

What are significant figures?

Steps only

Measured values have limited precision.

Answer

They are the meaningful digits in a measured value: all certain digits plus the first uncertain digit. They show the precision of the measurement.

Question 17

Water contains 15 ppm by mass CHCl₃. Find (i) percent by mass (ii) molality.

Steps only

Convert ppm to a mass fraction; multiply by 100 for percent.

Answer

15 ppm = 15/10⁶ = 1.5×10⁻⁵ mass fraction = 0.0015%. In 1 kg sample, CHCl₃=0.015 g=1.26×10⁻⁴ mol; water≈0.999985 kg. Molality≈1.26×10⁻⁴ mol kg⁻¹.

Question 18

Write in scientific notation: (i) 0.0048 (ii) 234000 (iii) 8008 (iv) 500.0 (v) 6.0012.

Steps only

Move the decimal so one non-zero digit remains to its left.

Answer

(i) 4.8×10⁻³
(ii) 2.34×10⁵
(iii) 8.008×10³
(iv) 5.000×10²
(v) 6.0012×10⁰.

Question 19

How many significant figures are in (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126000 (v) 500.0 (vi) 2.0034?

Steps only

Leading zeros do not count.

Answer

(i) 2
(ii) 3
(iii) 4
(iv) 3 (trailing zeros without a decimal are not certain)
(v) 4
(vi) 5.

Question 20

Round to three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808.

Steps only

Keep the first three significant digits.

Answer

(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2.81×10³.

Question 21

For N₂/O₂ data (14:16, 14:32, 28:32, 28:80), identify the law. Also convert (i) 1 km to mm and pm (ii) 1 mg to kg and ng (iii) 1 mL to L and dm³.

Steps only

Compare oxygen masses for the same nitrogen mass; they form simple ratios.

Answer

(a) For fixed nitrogen, oxygen masses 16 and 32 are in a simple whole-number ratio: law of multiple proportions. (b)

(i) 1 km=10⁶ mm=10¹⁵ pm.
(ii) 1 mg=10⁻⁶ kg=10⁶ ng.
(iii) 1 mL=10⁻³ L=10⁻³ dm³.

Question 22

Light travels at 3.0×10⁸ m s⁻¹. How far does it travel in 2.00 ns?

Steps only

Convert nanoseconds to seconds.

Answer

2.00 ns = 2.00×10⁻⁹ s. Distance = speed×time = 3.0×10⁸×2.00×10⁻⁹ = 0.60 m (2 significant figures).

Question 23

For A+B₂→AB₂, identify limiting reagent: (i) 300 A + 200 B₂ molecules (ii) 2 mol A+3 mol B₂ (iii) 100 A+100 B₂ (iv) 5 mol A+2.5 mol B₂ (v) 2.5 mol A+5 mol B₂.

Steps only

Balance: A + B₂ → AB₂, a 1:1 ratio.

Answer

The ratio A:B₂ required is 1:1.

(i) A limiting
(ii) A limiting
(iii) neither; exact amounts
(iv) B₂ limiting
(v) A limiting.

Question 24

N₂ + H₂ → NH₃: 2.00×10³ g N₂ reacts with 1.00×10³ g H₂. Find NH₃ formed and leftover reactant.

Steps only

Balance the equation, then convert reactants to moles.

Answer

Balance N₂+3H₂→2NH₃. n(N₂)=2000/28=71.4 mol; n(H₂)=1000/2=500 mol. N₂ needs 214 mol H₂, so N₂ limits. NH₃=142.8 mol×17=2.43×10³ g. H₂ left=(500−214)×2≈571 g.

Question 25

How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

Steps only

“mol” measures amount.

Answer

0.50 mol is an amount of substance. 0.50 M is a concentration: 0.50 mol Na₂CO₃ in each litre of solution.

Question 26

10 volumes H₂ react with 5 volumes O₂. What volume of water vapour forms?

Steps only

Use the balanced gaseous equation.

Answer

2H₂+O₂→2H₂O(g). The volume ratio H₂:O₂:H₂O is 2:1:2, so 10 volumes H₂ and 5 volumes O₂ produce 10 volumes water vapour (same temperature and pressure).

Question 27

Convert to basic SI units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg.

Steps only

1 pm=10⁻¹² m.

Answer

(i) 2.87×10⁻¹¹ m
(ii) 1.515×10⁻¹¹ m
(iii) 2.5365×10⁻² kg.

Question 28

Which has the most atoms: 1 g Au, 1 g Na, 1 g Li, or 1 g Cl₂?

Steps only

Compare atoms per gram: atomic moles per gram; for Cl₂ include two atoms per molecule.

Answer

1 g Li has the most atoms. For equal mass, the sample with the smallest molar mass has the most particles; Li has the lowest atomic molar mass. Cl₂ particles also contain two atoms, but its molar mass is much larger than Li’s.

Question 29

An ethanol–water solution has ethanol mole fraction 0.040. Find ethanol molarity assuming water density is 1 g mL⁻¹.

Steps only

Choose 1 mol total mixture; estimate its volume from total solution mass and density.

Answer

Take 1 mol mixture: ethanol=0.040 mol; water=0.960 mol, mass≈17.3 g. Volume≈17.3 mL=0.0173 L. M=0.040/0.0173≈2.31 mol L⁻¹.

Question 30

Find the mass in grams of one ¹²C atom.

Steps only

Use Avogadro’s constant: 6.022×10²³ atoms per mole.

Answer

One mole of ¹²C atoms has mass 12 g. Mass of one atom = 12/(6.022×10²³) = 1.99×10⁻²³ g.

Question 31

How many significant figures should the answers have? (i) 0.02856×298.15×0.112/0.5785 (ii) 5×5.364 (iii) 0.0125+0.7864+0.0215.

Steps only

For multiplication/division, use the fewest significant figures.

Answer

(i) 3 significant figures (the fewest is in 0.112).
(ii) 5×5.364 = 26.82, rounded to 30 (2×10¹), which has 1 significant figure.
(iii) 0.0125+0.7864+0.0215 = 0.8204, reported to four decimal places.

Question 32

Use isotope data: ³⁶Ar 35.96755 (0.337%), ³⁸Ar 37.96272 (0.063%), ⁴⁰Ar 39.9624 (99.600%).

Steps only

Convert percentages to decimal fractions.

Answer

Average mass = 35.96755(0.00337)+37.96272(0.00063)+39.9624(0.99600) ≈ 39.948 g mol⁻¹.

Question 33

How many atoms are in (i) 52 mol Ar (ii) 52 u He (iii) 52 g He?

Steps only

For moles, multiply by Avogadro’s constant.

Answer

(i) 52×6.022×10²³ = 3.13×10²⁵ atoms.
(ii) A He atom has mass about 4 u, so 52 u contains 13 atoms.
(iii) 52/4=13 mol He; atoms≈7.83×10²⁴.

Question 34

A C/H fuel gives 3.38 g CO₂ and 0.690 g H₂O on burning. 10.0 L at STP weighs 11.6 g. Find empirical formula, molar mass, and molecular formula.

Steps only

CO₂ gives the carbon amount; water gives twice its mole amount in hydrogen atoms.

Answer

n(C)=3.38/44.0=0.0768 mol. n(H)=2(0.690/18.0)=0.0767 mol. Ratio 1:1, empirical CH (13 g mol⁻¹). Gas molar mass=11.6×22.4/10.0≈26.0 g mol⁻¹. Multiplier≈2; molecular formula C₂H₂.

Question 35

What mass of CaCO₃ reacts completely with 25 mL of 0.75 M HCl?

Steps only

Balance equation and note 1 mol CaCO₃ reacts with 2 mol HCl.

Answer

CaCO₃+2HCl→CaCl₂+CO₂+H₂O. HCl moles=0.025×0.75=0.01875 mol. CaCO₃ moles=0.01875/2=0.009375 mol. Mass=0.009375×100.1≈0.938 g.

Question 36

How many grams HCl react with 5.0 g MnO₂?

Steps only

Use the balanced 1:4 mole ratio.

Answer

MnO₂+4HCl→MnCl₂+2H₂O+Cl₂. n(MnO₂)=5.0/86.94=0.0575 mol. HCl needed=4×0.0575=0.230 mol. Mass=0.230×36.46≈8.4 g.