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

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 exams must go through NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

NCERT Exercises

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

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.

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

Question 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.
Solution:
(i) 

Hence, 1 mole of C produces 44 g of CO₂
(ii)

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 \(\frac { 44 }{ 32 } \times 16=22g\quad of\quad { CO }_{ 2 }\)

(iii)

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

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:

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)

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

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

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

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

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

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

Question 24.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N_2(g) + H_2(g) → 2NH_3(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 + 3H_{3 →} 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.

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.

Question 27.
Convert the following into basic units :
(i) 28.7 pm
(ii) 15.15pm
(iii) 25365 mg
Solution.

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 Cl_2(g)
Solution.

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

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

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.

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

Solution.

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.

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.

Question 35.
Calcium carbonate reacts with aqueous HCI to give CaCI₂ and CO₂ according to the reaction,
CaC0_3(s) + 2HCI_(aq) —> CaCI_2(aq)+ CO_2(g) + H_2(s)O_(I). What mass of CaC03 is required to react completely with 25 mL of 0.75 M HCI?
Solution.

Question 36.
Chlorine is prepared in the laboratory by treating manganese dioxide (Mn02) with aqueous hydrochloric acid according to the reaction
4HCI_(aq) + MnO_2(s) → 2H₂O_(I) + MnCI_2(aq) + Cl_2(g)
How many grams of HCI react with 5.0 g of manganese dioxide?
Solution.

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NCERT Solutions for Class 11 Chemistry