Some Basic Concepts of Chemistry

Chapter 1 – Some Basic Concepts of Chemistry Exercise 25

Question 1.

Calculate the molecular mass of the following :

  • (i) H2o
  • (ii) CO2
  • (iii) CH4

Solution:

  • (i) Molecular mass of H2O :2 × 1 + 1 × 16 = 18u
  • (ii) Molecular mass of CO2 :1 × 12 + 2× 16 = 44 u
  • (iii) Molecular mass of CH4 : 12 + 4 × 1 = 16 u
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 1

Question 2.

Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).

Solution:

Molecular mass of Na2SO4 

= 2 × Atomic mass of Na + Atomic mass of S + 4 × Atomic mass of O

= 2 × 23 + 32 + 4 × 16

= 46 + 32 + 64

= 142 u.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 2

Question 3.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.
Solution:

Hence, the empirical formula is Fe2CO3

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 5
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 4

Question 4.

Calculate the amount of carbon dioxide that could be produced when

(1) 1 mole of carbon is burnt in air.

(2) 1 mole of carbon is burnt in 16 g of dioxygen.

(3) 2 moles of carbon are burnt in 16 g of dioxygen.

Solution:

  1.  Hence, 1 mole of C produces 44 g of CO2
  2. Hence, O2 is the limiting reagent.

∵ 32 g O2 reacts with C to produce 44 g of CO2

∵ 16 g O2 reacts with C to produce \frac { 44 }{ 32 } \times 16=22g\quad of\quad { CO }_{ 2 }

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 6

(3)

∵ 64 g O2 reacts with C to produce 88 g of CO2
∵ 16 g O2 reacts with C to produce \frac { 88 }{ 64 } \times 16 22g\quad of\quad { CO }_{ 2 }

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 7

Question 5.
Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.
Solution:

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 8

Chapter 1 – Some Basic Concepts of Chemistry Exercise 26

Question 6.
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 69 means that 69 g of HNO3 are dissolved in 100 g of the solution.

Question 7.
How much copper can be obtained from 100 g of copper sulphate (CuS04) ?
Solution:
Molar mass of CuSO4 

= 63.5 + 32 + 4 × 16

=63.5 +  32 + 64

= 159.5 amu or u
159.5 g of CuSO4 contains copper = 63.5 g
100 g of CuSO4 contains copper = \frac { 63.5 }{ 159.5 } \times 100  = 39.81 g

Question 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.
Solution.
For empirical formula, Fe2O3.
Molecular mass of Fe2O3 x 1

=  2 x 56 + 3 x 16

= 112 + 48

= 160
Molecular formula = u (Empirical formula)
∴ n=\frac { Molecular\quad mass }{ Empirical\quad formula\quad mass } =\frac { 160 }{ 160 } \Rightarrow n=1
∴ Molecular formula = (Fe2O3) x 1 = (Fe2O3)

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 9
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 10

Question 9.
Calculate the average atomic mass of chlorine using the following data.

Solution.

Question 10.

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.

Number of moles of carbon atoms in 3 moles of C2H6 = 3 x 2 = 6

(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms.

Number of moles of hydrogen atoms in 3 moles of C2H6 = 3 x 6 = 18

(iii) 1 mole of C2H6 = 6.022 x 1023 molecules Number of molecules in 3 moles of C2H6

= 3 x 6.022 x 1023 = 1.807 x 1024 molecules

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

Question 11.
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.
The molecular mass of sugar ((C12H22O11)

= 12 x 12 + 1 x 22 + 11 x 16

= 144 + 22 + 176

= 342

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 12
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 13

Question 12.
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.
Moles of methanol present in 2.5 L of 0.25 M solution

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 12
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 13


Question 13.
Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is 1 Pa = 1 Nm-2. If mass of air at sea level is 1034 g cm-2, calculate the pressure in pascal.
Solution.
Mass of air at sea level = 1034 g cm-2
Acceleration due to gravity, g = 9.8 m s-2
Pressure\quad of\quad air=\frac { 1034 }{ 1000 } \times 9.8\times 100\times 100=1.01332\times { 10 }^{ 5 }Pa

Question 14.
What is the S.I. unit of mass? How is it defined?
Solution.
The S I. unit of mass is kilogram. The amount of matter present in a substance is called mass. The unit of mass kilogram is defined as being equal to the mass of international prototype of the kilogram.

Question 15.

Match the following prefixes with their multiplies

Prefixes                 Multiples

  • (i) micro                     106
  • (ii) deca                     109
  • (iii) mega                  10-6
  • (iv) giga                     10-15
  • (v) femto                   10

Solution.

  • (i) micro → 10-6
  • (ii) deca → 10
  • (iii) mega → 106
  • (iv) giga → 109
  • (v) femto → 10-15

Question 16.
What do you mean by significant figures?
Solution.
Significant figures : The significant figures in a number are all the certain digits plus one doubtful digit, e.g., 2005 has four significant figures.

Question 17.

A sample of drinking water was found to be severely contaminated with chloroform, CHCI3, 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 million (106) parts

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 14
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 15

Question 18.

Express the following in the scientific notation:

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

Solution.

  • (i) 4.8 x 106
  • (ii) 2.34 x 105
  • (iii) 8.008 x 103
  • (iv) 5.00 x 102
  • (v) 6.0012 x 100 or 6.0012

Question 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

Solution.

  • (i) 2
  • (ii) 3
  • (iii) 4
  • (iv) 3
  • (v) 4
  • (vi) 5

Chapter 1 – Some Basic Concepts of Chemistry Exercise 27

Question 20.

Round up the following upto three significant figures:

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

Solution.

  • (i) 34.2
  • (ii) 10.4
  • (iii) 0.0460
  • (iv) 2810 or 2.81 x 103

Question 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                                                       16g
  • (ii) 14 g                                                     32 g
  • (iii) 28 g                                                   32 g
  • (iv) 28 g                                                   80 g

(a) Which law of chemical combination is obeyed bythe 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) Law of multiple proportions : This law was proposed by Dalton in 1803. It states that “If two elements can combine to form more than one compound the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

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 are in the ratio of 2 : 4 : 2 : 5 which is a simple whole number ratio. Hence, the given data obeys the law of multiple proportions.

(b)

  • (i) 106mm, 1015pm
  • (ii) 10-6 kg, 106ng
  • (iii) 10-3 L, 10-3 dm3

Question 22.
If the speed of light is 3.0 x 108 m s-1, calculate the distance covered by light in 2.00 ns.
Solution.
Speed of light = 3.0 x 108 m s-1
Distance covered by light in 2.00 ns = 3.0 x 108 x 2 x 10-9 = 6.00 x 10-1 m = 0.600 m

Question 23.

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 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

Solution.

According to the equation, one mole of A reacts with one mole of B and one atom of A reacts with one molecule of B.

  • (i) B is limiting reagent because 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be left in excess.
  • (ii) A
  • (iii) Both will react completely because it is stoichiometric mixture. No limiting reagent.
  • (iv) B
  • (v) A
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 16

Question 24.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2(g) + H2(g) → 2NH3(g)
(i) Calculate the mass of ammonia produced if 2.00 x 103 g dinitrogen reacts with 1.00 x 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.
The balanced chemical equation is N3 + 3H3 → 2NH3

Question 25.
How are 0.50 mol Na2CO3 and 0.50 M Na2CO3different?
Solution.
1 mol Na2CO3 \equiv 2 x 23 + 12 + 3 x 16 = 106 g mol-1
0.50 mol Na2CO3 \equiv 0.50 x 106= 53 g
0.50 M Na2CO3 solution means that 0.50 moles
or 53 g of Na2CO3 are dissolved in 1000 mL of solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 17

Question 26.
lf ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 18

Question 27.

Convert the following into basic units :

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

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 19

Chapter 1 – Some Basic Concepts of Chemistry Exercise 28

Question 28.

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.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 20

Question 29.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.
Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 21

Question 30.
What will be the mass of one 12C atom in g?
Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 22

Question 31.

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

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

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 23
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 24

Question 32.
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 25

Question 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.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 26

Question 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.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 27

Question 35.

Calcium carbonate reacts with aqueous HCI to give CaCI2 and CO2 according to the reaction,

CaC03(s) + 2HCI(aq) —> CaCI2(aq)+ CO2(g) + H2(s)O(I).

What mass of CaC03 is required to react completely with 25 mL of 0.75 M HCI?
Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 28

Question 36.

Chlorine is prepared in the laboratory by treating manganese dioxide (Mn02) with aqueous hydrochloric acid according to the reaction

4HCI(aq) + MnO2(s) → 2H2O(I) + MnCI2(aq) + Cl2(g)

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

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 28

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