**Class 11 NCERT solution of chapter 1- Some concepts of chemistry**

Here **class 11 chemistry NCERT solutions of chapter 1- Some basic concepts of chemistry** are explained for helping **class 11** students in their preparation of the exams. All the questions of **chapter 1 of class 11 chemistry** are explained by an expert of the subject by a step by step method so that every student could understand the **solutions** of each **NCERT question** of the **chemistry** textbook of class 11. These **NCERT solutions** will help you in clearing your concept of the basic laws of the **chemistry.**

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**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**Study notes of Maths and Science NCERT and CBSE from class 9 to 12**

**Class 11 NCERT solutions of Physics and Chemistry**

**Chapter 1 – Some concepts of the chemistry**

*Q1. Calculate the molar mass of the following.*

Ans.(i) **CH _{4}**

The molecular mass of carbon is = 12.011 u, the molecular mass of hydrogen = 1.oo8 u

The molecular mass of CH_{4}= 1 × 12.011 u + 4× 1.008 u = 12.011 u + 4.032 u = 16.043 u

(ii) **H _{2}O**

The molecular mass of hydrogen is = 1.008 u, the molecular mass of oxygen = 16.oo u

The molecular mass of H_{2}O= 2 × 1.008 u + 1× 16.00 u =2. 016 u + 16.00 u = 18.016 u

(iii) **CO _{2}**

The molecular mass of carbon is = 12.011 u, the molecular mass of oxygen = 16.oo u

The molecular mass of CO_{2}= 1× 12.011 u + 2× 16.00 u = 12.011 u + 32.00 u = 44.011 u

*Q2. Calculate the mass percent of different elements present in sodium sulphate ***(Na _{2}SO_{4}).**

Ans. Mass of Na in **(Na _{2}SO_{4}) = 2× 23.0 = 46 gm/mole of sodium sulphate**

Molar mass of **(Na _{2}SO_{4}) = 2 × 23.0 u + 1 × 32.066 u + 4 × 16.00 u = 46 u + 32.066 u + 64 u = 142.066 u**

32.379 ≈ 32.38 %

Mass of sulphur in **(Na _{2}SO_{4}) = 1 ×32.066 = 32.066 u**

22.57 ≈ 22.6 %

Mass of oxygen in **(Na _{2}SO_{4}) = 4× 23.0 = 92 gm/mole of sodium sulphate**

45.049 ≈ 45.05 %

**Q3.**Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1 % dioxygen by mass.

Ans. The given % of Fe = 69.9%

The atomic mass of Fe = 55.85 u

Relative no. of moles of Fe

The given % of O_{2 }= 30.1%

The atomic mass of Relative no. of moles of Fe16 u

Relative no. of moles of O_{2 }= 30.1%

The molar ratio between Fe and O

1.25: 1.88

Simple whole number molar ratio

2 : 3

Hence the empirical formula of an oxide of iron is Fe_{2}O_{3}

*Q4. 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.*

Ans. (i) *1 mole of carbon is burnt in air.*

The combustion equation of carbon is following

C(s) + O_{2} (g) ⇒ CO_{2}(g)

12 gm carbon (1 mole) reacts with 32 gm oxygen (1 mole) produces 44 gm of Carbon dioxide(1 mole)

Therefore 44 gm carbon dioxide is produced.

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

1 mole of carbon when reacts with 32 gm oxygen produces carbon di oxide= 44 gm

Therefore 1 mole of carbon when reacts with 16 gm oxygen produces carbon dioxide

Hence carbon dioxide produces = 22 gm

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

1 mole of carbon when reacts with 32 gm of dioxygen produces carbon dioxide = 44 gm

2 moles of carbon = 2 × 12 = 24 gm carbon

12 gm carbon corresponds to 32 gm related to the production of 44 gm carbon dioxide(1 mole)

16 gm of oxygen corresponds to the production of carbon dioxide = 0.5 mole

Since oxygen is limiting reactant here

So, 16 gm oxygen will react with limiting the amount of carbon = 12 × 0.5 = 6 gm carbon

Rest of carbon = 24 -6 = 18 gm will not react

Therefore carbon dioxide produces = 0.5 × 44 = 22 gm

**Q5. Calculate the mass of sodium acetate **(CH

_{3}COONa)

*required to make 500 mL of 0.375 molar aqueous solution. The molar mass of sodium acetate is 82.0245 g mol*^{–1}.Ans.** **0.375 M aqueous solution means that 1000 ml of the solution contains sodium acetate = 0.375 mole

No. of moles of sodium acetate (CH_{3}COONa) in 500 ml solution

The molar mass of sodium acetate = *82.0245 g mol ^{–1}*

m = 82.0245 × 0.1875 = 15.38 gm

Therefore required mass of sodium acetate is 15.38 gm

**Q6**.**Calculate the concentration of nitric acid in moles per litre in a sample that has a density of 1.41 g ****mL ^{-1}**

**and the mass per cent of nitric acid in it is being 69%.**

Ans. The given mass per cent of nitric acid is = 69 %

There is 69 gm of nitric acid in 100 gm solution of nitric acid

Molar mass of HNO_{3}) = 1 + 14 + 3 × 16 = 63 gm/mole

Mass = 69 gm

No of moles in 69 gm of nitric acid are = 1.095 mol

The density of nitric acid is = 1.41 gm/l

The volume of 100 gm nitric acid

= 70.92 ml

Now let’s find the concentration of nitric acid in moles /litre also known as the molarity of the solution

Conc. of nitric acid in moles /litre

**Q7. How much copper can be obtained from 100 g of copper sulphate (****CuSO _{4}**

**)**

**? (Atomic mass of Cu= 63.5 amu)**

Ans.

1 mole of CuSo_{4 }contains 1 mole (1 g atom) of Cu

Molar Mass of CuSo_{4 }= 63.5 + 32 +4 × 16 = 159.5 g mol^{-1}

Thus, Cu that can be obtained from 159.5 g of CuSo_{4 =} 63.5 g

Therefore Cu that can be obtained from 100 g of CuSo_{4}

**Q8.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 159.8 g** **mol**^{-1}**(Atomic mass: Fe = 55.85, O = 16.00 amu) Calculation of Empirical Formula. See Q3****.**

Ans. Empirical formula mass of Fe_{2}O_{3 }= 2 × 55.85 + 3 × 16.00 = 159.7 mol^{-1}

Hence, molecular formula is same as empirical formula, viz, Fe_{2}0_{3 }

**Q9.Calculate the atomic mass (average) of chlorine using the following data:**

% Natural Abundance | Molar Mass | |

75.77 | 34.9689 | |

24.23 | 36.9659 |

Ans.

Fractional abundance of ^{35}Cl = 0.7577,Molar mass = 34.9689

Average atomic mass of Cl

Therefore, average atomic mass =(0.7577) (34.9689 amu) + (0.2423) (36.9659 amu)

= 26.4959 + 8.9568 = 35.4527

Therefore the average atomic mass of chlorine =35.4527 u

**Q10. In three moles of ethane (****C _{2}H_{6}**

**), calculate the following:**

**(i) Number of moles of carbon atoms (ii) Number of moles of hydrogen atoms**

**(iii) Number of molecules of ethane**

Ans.(i)1 mole of C_{2}H_{6 }contains 2 moles of carbon atoms

3 moles of C_{2}H_{6 }will C-atoms =6 moles

(ii) 1 mole of C_{2}H_{6 }contains H-atoms = 18 moles

3 moles of C_{2}H_{6 }will contain H-atoms = 18 moles

(iii) In 1 mole of ethane the number of molecules of ethane = 6.023 × 10^{23}

In 3 moles of ethane the number of molecules of ethane = 3 × 6.023 × 10^{23}=18 .069 × 10^{23}

**Q11.What is the concentration of sugar (****C _{12}H_{22}O_{11}**

**) in mol**

**L**if its 20 g are dissolved in enough water to make a final volume up to 2 L?

^{-1 }Ans. No of moles of sugar

Mass of sugar = 20 gm

Molar mass of sugar(C_{12}H_{22}O_{11}) = 12 × 12 + 22 × 1 + 11 × 16 = 144 + 22 + 176 = 342 gm/mole

No. of moles of sugar

Molarity of sugar

Volume of solution = 2L

Hence **concentration of sugar (****C _{12}H_{22}O_{11}**

**) = 0.0293 mole/litre**

**Q12. 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?**

Ans. The density of methanol = 0.793 kg L^{-1}) = 793 gm L^{-1})

Molar mass of methanol (CH_{3}OH) = 12 + 3 + 16 + 1 = 32 gm

According to titration equation

Where and are the initial molarity and initial volume of the solution and and are the final molarity and volume of the solution.

= 24.78 mol/L, = ?, = 0.25 M = 0.25 mol/L, = 2L

24.78 × = 0.25 × 2.5

0.02522 L = 25.22 ml

Hence required volume to be added is = 25.22 ml

**Q13.Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal, is as shown below:****1 Pa = 1 ****Nm ^{-2}_{.}**

**If mass of air at sea level is 1034 g**

**cm**calculate the pressure in pascal.

^{-2},Ans. Pressure is defined as follows

_{ }

Where F is the force, A is the area, P is the pressure

Mass of air in per unit area is given = 1034 g cm^{-2})

So, A = 1 cm² = 10^{-4} m^{2}

m = 1034 gm = 1.034 kg, g = 9.8 m/s²

F = mg = 1.034 × 9.8 kg ms^{-1}= 10.1332 kg ms^{-1}

1 N = 1 kg m s^{-2}

1 Pa = 1 Nm^{-2}

1 Pa = 1 kg m^{-1}s^{-2}

Therefore the pressure of air in pascal = 1.01332 x 10^{-5} Pa

**Q14. What is the SI unit of mass? How is it defined?**

Ans. SI unit of mass is kilogram.

**Q15.** * Match the following prefixes with their multiples:*

Prefixes | Multiples |

Micro | 10^{6} |

Deca | 10^{9} |

Mega | 10^{-6} |

Giga | 10^{-15} |

Femto | 10 |

Ans.

Prefixes | Multiples |

Micro | 10^{-6} |

Deca | 10 |

Mega | 10^{6} |

Giga | 10^{9} |

Femto | 10^{-15} |

* *

**Q16. What do you mean by significant figures?**

Ans. The Significant figures are the meaningful digits which are known with certainty. Significant figures indicate uncertainty in the experimented value.

As an example suppose the result of an experiment is 23.5, total significant figures are 3

103.50, in this outcome the significant figures are 5

0.0056, in this outcome the significant figures are 2 because 0’s before 5 are not significant.

5600. has only 2 significant figure, trailing 0’s in a number without decimal are not significant.

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**NCERT Solutions for class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT Solutions for Class 11 Physics**

**chapter 3-Motion in a Straight Line**

**NCERT Solutions for Class 11 Chemistry**

**Chapter 1-Some basic concepts of chemistry**

**NCERT Solutions for Class 11 Biology**

**NCERT solutions for class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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