1. Some Basic Concepts of Chemistry

Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

Chapter 1. Some Basic Concepts of Chemistry

Class 11 Chemistry Chapter 1. Some Basic Concepts of Chemistry Exercise Questions and Answers :

1.1 Calculate the molar mass of the following:

(i) (ii) (iii)

Answer : (i) (water)

Molar mass of g/mol

(ii) (carbon dioxide)

Molar mass of g/mol

(iii) (methane)

Molar mass of g/mol

1.2 Calculate the mass per cent of different elements present in sodium sulphate ().

Answer: Atomic mass of Na = 23 g/mol

Atomic mass of S = 32 g/mol

Atomic mass of O = 16 g/mol.

Molar mass of g/mol

Calculate the mass percent of each element:

For Sodium (Na):

Mass percent of sodium (Na)

Sulphur (S):

Mass percent of sulphur (S)

Oxygen (O):

Mass percent of oxygen (O)

1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Answer: Atomic mass of Fe = 56 g/mol.

Atomic mass of O = 16 g/mol .

Moles of iron (Fe)

Moles of iron mole

Moles of oxygen (O)

Moles of oxygen (O) moles

Determine the mole ratio:

For iron:

For oxygen:

The mole ratio is 1 iron to 1.5 oxygen.

Iron: (multiple by 2 for whole number)

Oxygen: (multiple by 2 for whole number)

This gives a ratio of .

Therefore, the empirical formula of the iron oxide is .

1.4 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.

Answer: The balanced chemical equation for the combustion of carbon:

(i) When carbon (C) is burned in air, oxygen (O₂) from the air reacts with the carbon to form carbon dioxide (CO₂).

1 mole of carbon reacts with 1 mole of oxygen (from air) to form 1 mole of CO₂.

So, if we burn 1 mole of carbon in air, 1 mole of CO₂ is produced.

Amount of

Thus, 1 mole of CO₂ will be produced.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

Molar mass of oxygen (O₂) = 32 g/mol.

16 g of oxygen =

Therefore, 0.5 moles of CO₂ will be produced.

Amount of

(iii) 16 g of oxygen is 0.5 moles of O₂, which is less than the amount needed to burn 2 moles of carbon (because 2 moles of carbon would require 2 moles of oxygen).

So, oxygen is the limiting reagent here, and only 0.5 moles of carbon can react with 0.5 moles of oxygen. This will produce 0.5 moles of CO₂.

Amount of

1.5 Calculate the mass of sodium acetate ( ) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g .

Answer: Here, Molarity (M) = 0.375 mol/L

Volume (V) = 500 mL = 0.5 L (since 1 L = 1000 mL)

Molar mass of sodium acetate () = 82.0245 g/mol

We have,

The mass of sodium acetate:

We have,

The mass of sodium acetate required is approximately 15.38 grams.

1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g and the mass per cent of nitric acid in it being 69%.

Answer: Given, Density of the solution = 1.41 g/mL

Mass percent of nitric acid = 69%

Mass of nitric acid in 100 g solution = 69 grams (as 69% of 100 g is nitric acid) and the density is 1.41 g/mL

We have ,

Volume

The molar mass of nitric acid () = 63 g/mol.

Volume = 70.92 mL = 0.07092 L

We have,

mol/L

Therefore, the concentration of nitric acid in the given solution is approximately 15.44 moles per liter.

1.7 How much copper can be obtained from 100 g of copper sulphate ()?

Answer: Here, Atomic mass of Cu = 63.5 g/mol .

Atomic mass of S = 32 g/mol.

Atomic mass of O = 16 g/mol

Molar mass of

We have, moles

Therefore, from 100 g of copper sulfate, approximately 39.84 g of copper can be obtained.

1.8 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.

Answer: Given, Mass percent of iron (Fe) = 69.9%

Mass percent of oxygen (O) = 30.1%

Molar mass of iron (Fe) = 56.85 g/mol.

Molar mass of oxygen (O) = 16.00 g/mol.

We know that ,

moles

The mole ratio between iron and oxygen:

For iron:

For oxygen:

Iron: 1 × 2 = 2 (Multiplying by 2 for whole number)

Oxygen: 1.5 × 2 = 3 (Multiplying by 2 for whole number)

This gives a ratio of .

Therefore, the molecular formula of an oxide of iron is .

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

% Natural Abundance

Molar Mass

75.77

24.23

34.9689

36.9659

Answer: The average atomic mass :

Average Atomic Mass

The average atomic mass of chlorine is approximately 35.46 g/mol.

1.10 In three moles of ethane ( ), calculate the following:

(i) Number of moles of carbon atoms.

(ii) Number of moles of hydrogen atoms.

(iii) Number of molecules of ethane.

Answer: (i) Moles of carbon = 3 moles of ethane × 2 moles of carbon per mole of ethane

Moles of carbon = 6 moles of carbon

(ii) Moles of hydrogen = 3 moles of ethane × 6 moles of hydrogen per mole of ethane

Moles of hydrogen = 18 moles of hydrogen

(iii) 1 mole of any substance contains molecules.

Number of molecules of ethane ( )

Number of molecules of ethane molecules

1.11 What is the concentration of sugar ( ) in mol if its 20 g are dissolved in enough water to make a final volume up to 2L?

Answer: For the molar mass of sugar

Total molar mass

Given, mass

For the concentration (molarity): Volume of solution = 2 L

The concentration of sugar is 0.0292 mol/L.

1.12 If the density of methanol is 0.793 kg , what is its volume needed for making 2.5 L of its 0.25 M solution?

Answer: The molarity (M) of a solution is defined as:

We have,

The density of methanol is given as 0.793 kg/L .

1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:

1Pa = 1N

If mass of air at sea level is 1034 g , calculate the pressure in pascal.

Answer: The mass of air at sea level is .

We have,

Again,

We have,

1.14 What is the SI unit of mass? How is it defined?

Answer: The SI unit of mass is the kilogram (kg). It is defined as the mass of the international prototype of the kilogram, a platinum-iridium cylinder that was originally kept at the International Bureau of Weights and Measures (BIPM) in France. In 2019, the kilogram was redefined based on Planck’s constant.

1.15 Match the following prefixes with their multiples:

Prefixes

Multiple

(i) micro

(ii) deca

(iii) mega

(iv) giga

(v) femto

Answer: (i) micro – (ii) deca – (iii) mega – (iv) giga – (v) femto –

1.16 What do you mean by significant figures?

Answer: Significant figures are the meaningful digits in a measurement that include all certain digits plus one estimated or uncertain digit. They indicate the precision of a measurement. For example, in 9.4g, "4" is uncertain. The uncertainty is usually ±1 in the last digit, showing measurement limitations.

1.17 A sample of drinking water was found to be severely contaminated with chloroform, , supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(i) Express this in per cent by mass.

(ii) Determine the molality of chloroform in the water sample.

Answer: (i) Mass of chloroform

Mass of solution = 1000 g

(ii) Molar mass of

Given mass of chloroform

Moles of Chloroform

1.18 Express the following in the scientific notation:

(i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Answer: We have,

(i)

(ii)

(iii)

(iv)

(v)

1.19 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

Answer: The number of significant figures for each number:

(i) 0.0025 — 2 significant figures
(Zeros before the number are not significant; only 2 and 5 count.)

(ii) 208 — 3 significant figures
(There are no leading or trailing zeros in this case.)

(iii) 5005 — 4 significant figures
(Here, all the digits are significant, including the zero.)

(iv) 126,000 — 3 significant figures
(The trailing zeros are not considered significant unless a decimal point is shown.)

(v) 500.0 — 4 significant figures
(The decimal point makes the trailing zero significant.)

(vi) 2.0034 — 5 significant figures
(All digits, including the zero between the 3 and 4, are significant.)

1.20 Round up the following upto three significant figures:

(i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808

Answer: The numbers rounded to three significant figures:

(i) 34.216 → 34.2
(ii) 10.4107 → 10.4
(iii) 0.04597 → 0.0460 (The trailing zero is needed to keep three significant figures.)
(iv) 2808 → 2810

1.21 The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

Mass of dinitrogen

Mass of dioxygen

(i) 14 g

(ii) 14 g

(iii) 28 g

(iv) 28 g

16 g

32 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 = ......................

Answer: The given data obeys the Law of Multiple Proportions, which states:

"If two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a simple whole-number ratio."

For 14 g of N₂, the mass of O₂ is 16 g, 32 g

For 28 g of N₂, the mass of O₂ is 32 g, 80 g

For 14 g N₂:

For 28 g N₂:

These ratios (1:2, 2:5) are simple whole numbers, confirming the Law of Multiple Proportions.

(b) (i)

(ii)

(iii)

1.22 If the speed of light is , calculate the distance covered by light in 2.00 ns.

Answer: Here, and

We have,

So, the distance covered by light in 2.00 ns is 0.6 meters.

1.23 In a reaction,

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

Answer: Given, the balanced equation:

From the equation, 1 mole of A reacts with 1 mole of B₂ to form 1 mole of .

(i) 300 atoms of A + 200 molecules of B₂

Required ratio:

Given ratio:

Since 1 atom of A reacts with 1 molecule of , A is in excess.

Limiting reagent:

(ii)

Required ratio:

Given ratio:

A needs 2 mol of but 3 mol are available (excess ).

Limiting reagent: A

(iii) 100 atoms of A + 100 molecules of

Required ratio:

Given ratio:

Since both are present in the required ratio, no limiting reagent (reaction completes without excess).

Limiting reagent: None

(iv) 5 mol A + 2.5 mol B₂

Required ratio:

Given ratio:

Since 5 mol A requires 5 mol of , but only 2.5 mol is available, is the limiting reagent.

Limiting reagent:

(v) 2.5 mol A + 5 mol

Required ratio:

Given ratio:

Since 2.5 mol A requires 2.5 mol of , but 5 mol is available (excess), A is the limiting reagent.

Limiting reagent: A

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

(i) Calculate the mass of ammonia produced if g dinitrogen reacts with g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

Answer: Given: Mass of

Mass of

Reaction:

Molar Masses:

Molar mass

Molar mass of

(i) The mass of ammonia produced:

Required moles of

Moles of

Mass of :

Mass of

So, the mass of ammonia produced is approximately 2428.6 g.

(ii) Yes, since is the limiting reactant, will remain unreacted.

(iii) Moles of required for 71.43 moles of :

Required moles of

Moles of remaining

Mass of remaining

So, 571.42 g of will remain unreacted.

1.25 How are 0.50 mol and 0.50 M different?

Answer: 0.50 mol is the amount of sodium carbonate (0.50 moles), independent of the volume of solution. On the other hand, 0.50 M represents a molarity of sodium carbonate solution, meaning there are 0.50 moles of dissolved in 1 liter of solution. Molarity expresses concentration, while moles represent the quantity of substance.

1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer: The reaction between dihydrogen (H₂) and dioxygen (O₂) is:

Given that 10 volumes of react with 5 volumes of , the volume of produced will be:

Volume of volumes

So, 10 volumes of water vapour would be produced.

1.27 Convert the following into basic units:

(i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg

Answer: To convert the given values into basic units:

(i) 28.7 pm (picometers) to meters:

(ii) 15.15 pm to meters:

(iii) 25365 mg (milligrams) to kilograms:

1.28 Which one of the following will have the largest number of atoms?

(i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of (g)

Answer: (i) 1 g Au (s)

Molar mass of

Moles of

Number of atoms

(ii) 1 g Na (s)

Molar mass of

Moles of

Number of atoms

(iii) 1 g Li (s)

Molar mass of

Moles of

Number of atoms

(iv) 1 g of (g)

Molar mass of

Moles of

Number of atoms

The substance with the largest number of atoms is 1 g of Li (Lithium), with .

1.29 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).

Answer: Mole fraction of ethanol

Density of water

Moles of water

Volume of Solution

Let be the moles of ethanol.

The mole fraction of ethanol is given by:

Molarity (M) is given by:

The molarity of the ethanol solution is 2.32 M.

1.30 What will be the mass of one atom in g?

Answer: The atomic mass of one mole of is 12 g.

Avogadro’s number ,

Mass of one

1.31 How many significant figures should be present in the answer of the following calculations?

(i)

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

Answer: (i)

0.02856 has 4 significant figures.

298.15 has 5 significant figures.

0.112 has 3 significant figures.

0.5785 has 4 significant figures.

Since we use multiplication and division, the final answer should have 3 significant figures (since 0.112 has the least number of significant figures).

(ii)

5 is an ambiguous case; if it is considered an exact number (like a count), it has infinite significant figures.

5.364 has 4 significant figures.

If the 5 is considered to have only 1 significant figure, then the result should have 1 significant figure.

(iii)

0.0125 has 4 decimal places.

0.7864 has 4 decimal places.

0.0215 has 4 decimal places.

Since all terms have 4 decimal places, the result should also have 4 decimal places.

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

Isotope

Isotopic molar mass

Abundance

35.96755 g

37.96272 g

39.9624 g

0.337%

0.063%

99.600%

Answer: We know that,

1.33 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.

Answer: (i) 52 moles of Ar (Argon) :

Avogadro’s number

So, the number of atoms in 52 moles of Argon is:

Number of atoms

Thus, the number of atoms in 52 moles of Argon is .

(ii) 52 u of He (Helium)

The mass of one helium atom is 4 u.

Therefore, the number of atoms in 52 u is:

Number of atoms

Thus, the number of atoms in 52 u of He is 13 atoms.

(iii) 52 g of He (Helium)

The number of moles of helium in 52 grams is:

Number of moles

Number of atoms

Thus, the number of atoms in 52 g of He is .

1.34 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.

Answer: (i) Molar mass of

Mass of carbon in 3.38 g of :

Mass of

Molar mass of

Mass of

The fuel gas contains only carbon and hydrogen, so the total mass of the gas sample

Moles of Carbon

Moles of Hydrogen

Since the ratio of , the empirical formula is .

(ii) Mass of 10.0 L of gas at

At STP, 1 mole of gas occupies 22.4 L.

Molar mass: Molar mass

(iii) Empirical formula mass of

Molecular formula factor:

Thus, the molecular formula is (acetylene).

1.35 Calcium carbonate reacts with aqueous HCl to give and according to the reaction,

What mass of is required to react completely with 25 mL of 0.75 M HCl ?

Answer: The given balanced chemical equation is:

Volume of

Molarity of (moles of HCl per liter)

Therefore, the moles of CaCO₃ required to react with 0.01875 moles of HCl is:

Moles of

The molar mass of is:

; and

Molar mass of

Mass of

The mass of required to react completely with 25 mL of is 0.937 g.

1.36 Chlorine is prepared in the laboratory by treating manganese dioxide () with aqueous hydrochloric acid according to the reaction

How many grams of HCl react with 5.0 g of manganese dioxide?

Answer: The balanced equation is:

Molar mass of :

Manganese ; Oxygen

Molar mass of

The number of moles of :

Moles of

Molar mass of

Mass of

The mass of HCl required to react with 5.0 g of manganese dioxide is 8.39 g.