## Class 11 Chemistry Chapter 1 NCERT Solutions FREE PDF Download

Embark your journey with Vedantu through class 11 chemistry chapter 1 NCERT solutions. This chapter delves into "Some Basic Concepts of Chemistry," elucidating fundamental principles crucial for understanding the subject's intricacies. By accessing the class 11 chemistry chapter 1 PDF, students gain comprehensive insights into the core concepts outlined in the curriculum. With a focus on clarity and depth, these resources serve as indispensable tools for students navigating through chemistry class 11, chapter 1 of their Chemistry syllabus.

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## Quick Insights of “Some Basic Concepts of Chemistry” Class 11 NCERT Solutions

NCERT Solutions for Class 11 Chemistry Chapter 1 will give you insights about the General Introduction: Importance and Scope of Chemistry.

The section will give you crisp learnings on the Nature of matter, laws of chemical combination, Dalton's atomic theory and also the concept of elements, atoms, and molecules.

The understanding related to topics like Atomic and molecular masses, mole concept and molar mass, percentage composition, and empirical and molecular formulas will help the students to solve numerical-based problems.

Detailed explanations are given on chemical reactions, stoichiometry, and calculations based on stoichiometry.

Using these class 11 chemistry chapter 1 NCERT solutions can help students analyse their level of preparation and understanding of concepts.

Chemistry class 11 chapter 1 topics are included according to the revised syllabus for the academic year 2024-25.

provides resources such as class notes, important concepts, and formulas exemplar solutions.

## Access NCERT Solutions for Class 11 Chemistry Chapter 1- Some basic concepts of Chemistry

1. Calculate the molecular mass of the following:

i) ${{H}_{2}}O$

Ans: ${{H}_{2}}O$

The molecular mass of water, ${{\text{H}}_{\text{2}}}\text{O}$, can be calculated by the following steps given below:

= (2 $\times $ Atomic mass of hydrogen element) + (1 $\times $ Atomic mass of oxygen element)

= $\text{(2 }\!\!\times\!\!\text{ 1}\text{.0084 u) + (1 }\!\!\times\!\!\text{ 16}\text{.00 u)}$

= 2.016 u + 16.00 u

= 18.016 u or 18.02 u

ii) $C{{O}_{2}}$

Ans: $\text{C}{{\text{O}}_{\text{2}}}$

The molecular mass of carbon dioxide,$\text{C}{{\text{O}}_{\text{2}}}$, is calculated below:

= (1 $\times $ Atomic mass of carbon) + (2 $\times $ Atomic mass of oxygen)

= $\text{(1 }\!\!\times\!\!\text{ 12}\text{.011 u) + (2 }\!\!\times\!\!\text{ 16}\text{.00 u)}$

= 12.011 u + 32.00 u

= 44.01 u

iii) $C{{H}_{4}}$

Ans: $\text{C}{{\text{H}}_{\text{4}}}$

The molecular mass of methane,$\text{C}{{\text{H}}_{\text{4}}}$, is calculated below in step by step manner:

= (1 $\times $ Atomic mass of carbon) + (4 $\times $ Atomic mass of hydrogen)

= $\text{(1 }\!\!\times\!\!\text{ 12}\text{.011 u) + (4 }\!\!\times\!\!\text{ 1}\text{.008 u)}$

= 12.011 u +4.032 u

= 16.043 u

2. Calculate the mass percent of different elements present in sodium sulphate$\left( \text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} \right)$ .

Ans: The given compound in the question is sodium sulphate and its formula is $\text{N}{{\text{a}}_{2}}S{{O}_{4}}$. Its molecular formula is calculated below:

$\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{=}\left[ \left( \text{2 }\!\!\times\!\!\text{ 23}\text{.0} \right)\text{+}\left( \text{32}\text{.066} \right)\text{+4}\left( \text{16}\text{.00} \right) \right]$

= 142.066 g

Now, let us find the mass percentage of each element in the given compound using the formula given below:

Mass percent of an element $=\dfrac{\text{Mass of that element in the compound }}{\text{Molar mass of the compound}}\times 100$

$\therefore $Mass percent of sodium:

$\text{=}\dfrac{\text{46}\text{.0 g}}{\text{142}\text{.066 g}}\text{ }\!\!\times\!\!\text{ 100}$

= 32.379

= 32.4%

Now, let us find the mass percentage of sulphur:

$\text{=}\dfrac{\text{32}\text{.066 g}}{\text{142}\text{.066 g}}\text{ }\!\!\times\!\!\text{ 100}$

= 22.57

= 22.6%

Now, the mass percentage of oxygen:

$\text{=}\dfrac{\text{64}\text{.0 g}}{\text{142}\text{.066 g}}\text{ }\!\!\times\!\!\text{ 100}$

= 45.049

= 45.05%

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

Ans: We are given that the percentage of iron by mass is 69.9% and the percentage of oxygen by mass is 30.1%.

Now, we can calculate the relative moles of iron by using the formula given below:

$=\dfrac{\%\text{ of iron by mass}}{\text{Atomic mass of iron}}$

$=\dfrac{69.9}{55.85}$

= 1.25

In the same way we can calculate the relative moles of oxygen:

$=\dfrac{\%\text{ of oxygen by mass}}{\text{Atomic mass of oxygen}}$

$=\dfrac{30.1}{16.00}$

= 1.88

Since we have relative moles of both elements, we can calculate the simpler molar ratio.

= 1.25: 1.88

= 1: 1.5

= 2: 3

So, now we can write the empirical formula of iron oxide as $F{{e}_{2}}{{O}_{3}}$.

4. Calculate the amount of carbon dioxide that could be produced when

i) 1 mole of carbon is burnt in the air.

Ans: It is possible to express the balanced response of carbon combustion as follows:

$\text{C+}{{\text{O}}_{\text{2}}}\to \text{C}{{\text{O}}_{\text{2}}}$

Carbon dioxide is produced when one mole of carbon is burned in one mole of dioxygen (air) using the balancing equation.

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

Ans: It is possible to express the balanced response of carbon combustion as follows:

$\text{C+}{{\text{O}}_{\text{2}}}\to \text{C}{{\text{O}}_{\text{2}}}$

There is only 16 g of dioxygen accessible, as stated in the question. In this case, it will react with 0.5 moles of carbon to produce 22 g of $C{{O}_{2}}$. In this sense, it is a limiter.

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

Ans: It is possible to express the balanced response of carbon combustion as follows:

$\text{C+}{{\text{O}}_{\text{2}}}\to \text{C}{{\text{O}}_{\text{2}}}$

It appears that just 16 g of dioxygen is accessible. In other words, it's a reaction-suppressing agent. This means that with just half as much carbon as dioxygen, 22 g of carbon dioxide may be produced.

5. Calculate the mass of sodium acetate $\text{(C}{{\text{H}}_{\text{3}}}\text{COONa)}$ required to make 500 mLof 0.375 molar aqueous Ans. Molar mass of sodium acetate is$\text{82}\text{.0245 g mo}{{\text{l}}^{-1}}$.

Ans: 0.375 M aqueous Ans of sodium acetate means 1000 ml of Ans contains 0.375 moles of sodium acetate.

So, we can calculate the number of moles of sodium acetate in 500 ml. This is given below:

$=\dfrac{0.375}{1000}\times 500$

= 0.1875 mole

We are also given the molar mass of sodium acetate is $82.0245\text{ g mol}{{\text{e}}^{-1}}$.

Required mass of sodium acetate can be calculated as:$=\left( 82.0245\text{ g mo}{{\text{l}}^{-1}} \right)\left( 0.1875\text{ mole} \right)$

= 15.38 g

6. Calculate the concentration of nitric acid in moles per liter in a sample which has a density, $1.41\text{ g m}{{\text{L}}^{-1}}$ and the mass percent of nitric acid in it being 69%.

Ans: We are given the mass percentage of nitric acid in the sample as 69%.

So, we can say that 100 g of nitric acid contains 69 g of nitric acid by mass.

Molar mass of nitric acid $\text{(HN}{{\text{O}}_{\text{3}}}\text{)}$ is calculated as below:

= $\text{1 + 14 + (3 }\!\!\times\!\!\text{ 16)}$

= 1 + 14 + 48

= $63\text{ g mo}{{\text{l}}^{-1}}$

As we have the molar mass of nitric acid so, we can find the number of moles in 69 g of nitric acid as given below:

$=\dfrac{69\text{ g}}{63\text{ g mo}{{\text{l}}^{-1}}}$

= 1.095 mol

Volume of 100 g of nitric acid And can be calculated as:

$=\dfrac{\text{Mass of solution}}{\text{density of solution}}$

$=\dfrac{100\text{ g}}{1.41\text{ g m}{{\text{L}}^{-1}}}$

$\text{=70}\text{.92 mL}$

$\text{=70}\text{.92 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{ L}$

Now, we can calculate the concentration of nitric acid as:

$\text{=}\dfrac{\text{1}\text{.095 mole}}{\text{70}\text{.92 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{ L}}$

= 15.44 mol/ L

Therefore, the concentration of nitric acid is 15.44 mol/ L

7. How much copper can be obtained from 100 g of copper sulphate$\text{(CuS}{{\text{O}}_{\text{4}}}\text{)}$?

Ans: In copper sulfate, we can see that there is one atom of copper so, we can say that 1 mole of $CuS{{O}_{4}}$ will have 1 mole of copper.

The molar mass of copper sulphate is calculated below:

$CuS{{O}_{4}}$ = $\text{63}\text{.5 + 32 + (4 }\!\!\times\!\!\text{ 16)}$

= 63.5 + 32.0 + 64.0

= 159.5 g

We can say that 159.5 g of $CuS{{O}_{4}}$ will have 63.5 g of copper.

$\Rightarrow 100\text{ g of CuS}{{\text{O}}_{4}}$ will contain $\dfrac{\text{63}\text{.5 }\!\!\times\!\!\text{ 100 g}}{\text{159}\text{.5}}$ of copper.

So, the amount of copper that can be obtained from 100 g of $\text{CuS}{{\text{O}}_{\text{4}}}$$=$$\dfrac{\text{63}\text{.5 }\!\!\times\!\!\text{ 100 g}}{\text{159}\text{.5}}$

= 39.81 g

8. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is $\text{159}\text{.69 g mo}{{\text{l}}^{\text{-1}}}$

Ans: We are given the mass percentage of iron (Fe) as 69.9% and the percentage of oxygen (O) as 30.1%.

We know the atomic mass of iron is 55.85.

The number of moles of iron present in the oxide will be $=\dfrac{69.90}{55.85}$

$=1.25$

The atomic mass of oxygen is 16.

The number of moles of oxygen present in the oxide will be $=\dfrac{30.1}{16.0}$

$=1.88$

The ratio of iron to the oxygen in the oxide is calculated below:

= 1.25: 1.88

$=\dfrac{1.25}{1.25}:\text{ }\dfrac{1.88}{1.25}$

= 1: 1.5

= 2: 3

According to the ratio the formula of the oxide will be $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$.

Empirical formula mass of $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{ = }\!\![\!\!\text{ (2 }\!\!\times\!\!\text{ 55}\text{.85) + (3 }\!\!\times\!\!\text{ 16}\text{.00) }\!\!]\!\!\text{ = 159}\text{.7}$ g/ mol

Given molar mass of$\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ = 159.69 g/ mol

Multiplying the empirical formula by n gives the molecular formula for the molecule. This is calculated below:

$\text{n = }\dfrac{\text{Molar mass}}{\text{Empirical formula mass}}\text{=}\dfrac{\text{159}\text{.69 g}}{\text{159}\text{.7 g}}$

= 0.999 = 1

So, the empirical formula of the oxide is $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and the value of n is 1. Therefore, the molecular formula of the oxide is $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$.

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

% Natural Abundance | Molar Mass | |

$^{\text{35}}\text{Cl}$ | 75.77 | 34.9689 |

$^{\text{37}}\text{Cl}$ | 24.23 | 36.9659 |

Ans:The average atomic mass of chlorine is calculated below:

$\text{=}\left[ \left( \text{Fractional abundance of}{{\text{ }}^{\text{35}}}\text{Cl} \right)\left( \text{Molar mass of}{{\text{ }}^{\text{35}}}\text{Cl} \right)\text{+}\left( \text{Fractional abundance of}{{\text{ }}^{\text{37}}}\text{Cl} \right)\left( \text{Molar mass of }{{\text{ }}^{\text{37}}}\text{Cl} \right) \right]$$\text{=}\left[ \left\{ \left( \dfrac{\text{75}\text{.77}}{\text{100}} \right)\left( \text{34}\text{.9689u} \right) \right\}\text{+}\left\{ \left( \dfrac{\text{24}\text{.23}}{\text{100}} \right)\left( \text{36}\text{.9659u} \right) \right\} \right]$

= 26.4959 + 8.9568

= 35.4527 u

So, the average atomic mass of chlorine is 35.4527 u.

10. In three moles of ethane$\text{(}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}\text{)}$, calculate the following:

i). Number of moles of carbon atoms.

Ans: 2 Moles of carbon atoms are present in each mole of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

Therefore, number of moles of carbon atoms in 3 moles of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

= $\text{2 }\!\!\times\!\!\text{ 3 = 6}$

ii) Number of moles of hydrogen atoms.

Ans: 6 moles of hydrogen atoms are present in each mole of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

Therefore, number of moles of carbon atoms in 3 moles of${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

= $\text{3 }\!\!\times\!\!\text{ 6 = 18}$

iii) Number of molecules of ethane.

Ans: $6.023\text{ }\times \text{ 1}{{\text{0}}^{23}}$ molecules of ethane are present in each mole of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

Therefore, number of molecules in 3 moles of${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

$=3\times 6.023\times {{10}^{23}}=18.069\times {{10}^{23}}$

11. What is the concentration of sugar $\left( {{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}} \right)$in $\text{mol }{{\text{L}}^{\text{-1}}}$if its 20 g are dissolved in enough water to make a final volume up to 2 L?

Ans: Molarity (M) of the Ans can be calculated by the formula given below:

$=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution in Litres}}$

This can be further written as

$=\dfrac{\text{Mass of sugar / molar mass of sugar}}{\text{2 L}}$

Putting the values, we get:

$\text{=}\dfrac{\text{20 g/}\left[ \left( \text{12 }\!\!\times\!\!\text{ 12} \right)\text{+}\left( \text{1 }\!\!\times\!\!\text{ 22} \right)\text{+}\left( \text{11 }\!\!\times\!\!\text{ 16} \right) \right]\text{ g}}{\text{2 L}}$

$\text{=}\dfrac{\text{20 g/ 342 g}}{\text{2 L}}$

$\text{=}\dfrac{\text{0}\text{.0585mol}}{\text{2L}}$

$\text{=0}\text{.02925 mol }{{\text{L}}^{\text{-1}}}$

So, the molar concentration of sugar is $\text{=0}\text{.02925 mol }{{\text{L}}^{\text{-1}}}$.

12. If the density of methanol is $\text{0}\text{.793 kg }{{\text{L}}^{\text{-1}}}$, what is its volume needed for making 2.5 L of its 0.25 M Ans?

Ans: The molar mass of methanol is calculated below: $\left( \text{C}{{\text{H}}_{3}}\text{OH} \right)=\left( 1\times 12 \right)+\left( 4\times 1 \right)+\left( 1\times 16 \right)$

$\text{=32 g mo}{{\text{l}}^{\text{-1}}}$

$\text{=0}\text{.032 kg mo}{{\text{l}}^{\text{-1}}}$

Molarity of the methanol will be $\text{=}\dfrac{\text{0}\text{.793 kg }{{\text{L}}^{\text{-1}}}}{\text{0}\text{.032 kg mo}{{\text{l}}^{\text{-1}}}}$

$=24.78\text{ mol }{{\text{L}}^{-1}}$

(Since density is mass per unit volume)

Now, to find the volume we have to write the formula,

${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{=}{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$

1 is for the given Ans and 2 is for the Ans to be prepared.

Putting the values, we get:

$\left( 24.78\text{ mol }{{\text{L}}^{-1}} \right){{\text{V}}_{1}}=\left( 2.5\text{L} \right)\left( 0.25\text{ mol }{{\text{L}}^{-1}} \right)$

${{\text{V}}_{1}}=0.0252\text{ L}$

${{\text{V}}_{1}}=25.22\text{ mL}$

13. Pressure is determined as force per unit area of surface. The SI unit of pressure,Pascal is as shown below:

$\text{1 Pa=1 N}{{\text{m}}^{\text{-2}}}$

If the mass of air at sea level is $\text{1034 g c}{{\text{m}}^{\text{-2}}}$, calculate the pressure in Pascal.

Ans: Force per unit area of a surface is described as pressure. This is calculated below:

$\text{P=}\dfrac{\text{F}}{\text{A}}$

$\text{=}\dfrac{\text{1034g }\!\!\times\!\!\text{ 9}\text{.8 m}{{\text{s}}^{\text{-2}}}}{\text{c}{{\text{m}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1 kg}}{\text{1000 g}}\text{ }\!\!\times\!\!\text{ }\dfrac{{{\left( \text{100} \right)}^{\text{2}}}\text{ c}{{\text{m}}^{\text{2}}}}{\text{1}{{\text{m}}^{\text{2}}}}$

$=1.01332\times {{10}^{5}}\text{ kg }{{\text{m}}^{-1}}\text{ }{{\text{s}}^{-2}}$

We know the relation of force can be written as:

$\text{1 N=1 kg m }{{\text{s}}^{\text{-2}}}$

Then we can write,

$\text{1 Pa=1 N}{{\text{m}}^{\text{-2}}}\text{=1 kg }{{\text{m}}^{\text{-2}}}\text{ }{{\text{s}}^{\text{-2}}}$

$\text{1 Pa=1 kg }{{\text{m}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-2}}}$

$\text{Pressure=1}\text{.01332 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{ Pa}$

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

Ans: The kilogram is the SI unit of mass in the SI system (kg). An international kilogram prototype is defined as one kilogram.

15. Match the following prefixes with their multiples:

Prefixes | Multiples | |

Micro | $\text{1}{{\text{0}}^{\text{6}}}$ | |

Deca | $\text{1}{{\text{0}}^{\text{9}}}$ | |

Mega | $\text{1}{{\text{0}}^{\text{-6}}}$ | |

Giga | $\text{1}{{\text{0}}^{\text{-15}}}$ | |

Femto | 10 |

Ans: The multiples are matched with their prefixes in the table given below:

Prefixes | Multiples | |

Micro | $\text{1}{{\text{0}}^{\text{-6}}}$ | |

Deca | 10 | |

Mega | $\text{1}{{\text{0}}^{\text{6}}}$ | |

Giga | $\text{1}{{\text{0}}^{\text{9}}}$ | |

Femto | $\text{1}{{\text{0}}^{\text{-15}}}$ |

16. What do you mean by significant figures?

Ans: Significant figures are those digits that have meaning and are recognized to be assured of their value.

They are used to show uncertainty in an experiment or a computed number.

For example, if 18.2 mL is the result of an experiment, then 18 is certain while 2 is uncertain, and the total number of significant figures is 3.

Since the final digit signifies uncertainty, significant figures are defined as the sum of all of a number's decimal places.

17. A sample of drinking water was found to be severely contaminated with chloroform, $\text{CHC}{{\text{l}}_{\text{3}}}$, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

i) Express this in percent by mass.

Ans: 1 part out of 1 million $\left( \text{1}{{\text{0}}^{\text{6}}} \right)$parts is equal to 1 ppm.

Therefore, mass percent of $15$ ppm chloroform in water will be

$\text{=}\dfrac{\text{15}}{\text{1}{{\text{0}}^{\text{6}}}}\text{ }\!\!\times\!\!\text{ 100}$

$\simeq 1.5\times {{10}^{-3}}%$

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

Ans: 100 g of the sample will have $1.5\text{ }\times \text{ 1}{{\text{0}}^{-3}}$of $\text{CHC}{{\text{l}}_{\text{3}}}$.

So 1000 g of the sample contains will have $1.5\text{ }\times \text{ 1}{{\text{0}}^{-2}}$of $\text{CHC}{{\text{l}}_{\text{3}}}$ .

Therefore, molality of chloroform in water is calculated as:

$\text{=}\dfrac{\text{1}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{ g}}{\text{Molar mass of CHC}{{\text{l}}_{\text{3}}}}$

For we want molar mass of chloroform.

$\text{Molar mass of CHC}{{\text{l}}_{\text{3}}}\text{=12}\text{.00+1}\text{.00+3}\left( \text{35}\text{.5} \right)\text{ = 119}\text{.5 g mo}{{\text{l}}^{\text{-1}}}$

$\therefore $ Molality of chloroform in water$\text{=0}\text{.0125 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{m}$

$\text{=1}\text{.25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{m}$

18. Express the following in the scientific notation:

Ans: If we want to express the numbers in scientific notation, we must first put the number in decimal form and multiply it by 10 with some power. These are given below:

0.0048

Ans: $0.0048=4.8\times {{10}^{-3}}$

234,000

Ans: $234,000=2.34\times {{10}^{5}}$

8008

Ans: $8008=8.008\times {{10}^{3}}$

500.0

Ans: $500.0=5.000\times {{10}^{2}}$

6.0012

Ans: $6.0012=6.0012\times {{10}^{0}}$

19. How many significant figures are present in the following?

Ans: There are some rules to find the number of significant figures and by following the rules the significant figures are given below:

0.0025

Ans: There are 2 significant figures.

208

Ans: There are 3 significant figures.

5005

Ans: There are 4 significant figures.

126,000

Ans: There are 3 significant figures.

500.0

Ans: There are 4 significant figures.

2.0034

Ans: There are 5 significant figures.

20. Round up the following up to three significant figures.

Ans: These can be written as:

34.216

Ans: 34.216 = 34.2

10.4107

Ans: 10.4107 = 10.4

0.04597

Ans: 0.04597 = 0.0460

2808

Ans: 2808 = 2810

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

Mass of dinitrogen | Mass of dioxygen | |

14 g | 16 g | |

14 g | 32 g | |

28 g | 32 g | |

28 g | 80 g |

a) Which law of chemical combination is obeyed by the above experimental data? Give it a statement.

Ans: The masses of dioxygen that will combine with the fixed dinitrogen mass are 32 g, 64 g, 32 g, and 80 g, respectively. The mass ratios of dioxygen are 1: 2: 1: 5. In light of this fact, the supplied experimental data is in accordance with the law of multiple proportions. One element's mass multiplied by the fixed mass of another element must be small whole numbers if two elements combine to produce more than one compound, according to this rule.

b) Fill in the blanks in the following conversions:

Ans: All the conversions are given below:

i) 1 km = …………….. mm …………..pm

Ans: $\text{1 km = 1 km }\!\!\times\!\!\text{ }\dfrac{\text{1000 m}}{\text{1 km}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{100 cm}}{\text{1 m}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{10 mm}}{\text{1 cm}}$

$\text{1km = 1}{{\text{0}}^{\text{6}}}\text{mm}$

$\text{1 km = 1 km }\!\!\times\!\!\text{ }\dfrac{\text{1000 m}}{\text{1 km}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1 pm}}{\text{1}{{\text{0}}^{\text{-12}}}\text{ m}}$

$\text{1km=1}{{\text{0}}^{\text{15}}}\text{pm}$

$\text{Hence, 1 km = 1}{{\text{0}}^{\text{6}}}\text{ mm = 1}{{\text{0}}^{\text{15}}}\text{ pm}$

ii) 1 mg = ……………... kg …………… ng

Ans: $\text{1 mg = 1 mg }\!\!\times\!\!\text{ }\dfrac{\text{1 g}}{\text{1000 mg}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1 kg}}{\text{1000 g}}$

$\text{1mg = 1}{{\text{0}}^{\text{6}}}\text{ kg}$

$\text{Hence, 1mg = 1}{{\text{0}}^{\text{-6}}}\text{ kg = 1}{{\text{0}}^{\text{6}}}\text{ ng}$

iii) 1 mL = ……………. L ……………… $\text{d}{{\text{m}}^{\text{3}}}$

Ans: $\text{1 mL = 1 mL }\!\!\times\!\!\text{ }\dfrac{\text{1 L}}{\text{1000 mL}}$

$\text{1 mg = 1}{{\text{0}}^{\text{-3}}}\text{ L}$

$\text{1 mL = 1 c}{{\text{m}}^{\text{3}}}\text{ = 1 c}{{\text{m}}^{\text{3}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1 dm }\!\!\times\!\!\text{ 1 dm }\!\!\times\!\!\text{ 1 dm}}{\text{10 cm }\!\!\times\!\!\text{ 10 cm }\!\!\times\!\!\text{ 10 cm}}$

$\text{1 mL = 1}{{\text{0}}^{\text{-3}}}\text{ d}{{\text{m}}^{\text{3}}}$

$\text{Hence, 1 mL = 1}{{\text{0}}^{\text{-3}}}\text{ L = 1}{{\text{0}}^{\text{-3}}}\text{ d}{{\text{m}}^{\text{3}}}$

22. If the speed of light is $\text{3}\text{.0 }\times \text{ 1}{{\text{0}}^{\text{8}}}\text{ m}{{\text{s}}^{\text{-1}}}$, calculate the distance covered by light in 2.00 ns.

Ans: From the question we can see that the time taken to cover the distance is 2.00ns. This can be written as:

$\text{=2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-9}}}\text{s}$

We know the speed of light$\text{= 3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{8}}}\text{ m}{{\text{s}}^{\text{-1}}}$

So, the distance travelled by light in 2.00ns will be:

$=\text{Speed of light }\times \text{ Time taken}$

$\text{=}\left( \text{3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{8}}}\text{m}{{\text{s}}^{\text{-1}}} \right)\left( \text{2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-9}}}\text{s} \right)$

$\text{=6}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-1}}}\text{m}$

= 0.600 m

23. In a reaction

$\text{A+}{{\text{B}}_{\text{2}}}\to \text{A}{{\text{B}}_{\text{2}}}$

Identify the limiting reagent, if any in the following reaction mixtures.

Ans: Reactants that serve as limiting agents restrict the amount of a response. An initial reaction product is consumed before any more products are created, which stops the reaction and limits the quantity of products that are formed.

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

Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. In other words, 200 molecules of B will react with 200 atoms of A, leaving 100 atoms of A unusable as a result. B is, thus, the limiting reagent in the reaction.

(ii) 2 mol A + 3 mol B

Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. In other words, 2 moles of A will react with just 2 moles of B. Consequently, 1 mol of A will not be used up in the process. This means that A is considered to be the limiting reagent.

(iii) 100 atoms of A + 100 molecules of B

Ans: According to the given reaction, 1 atom of A reacts with 1 molecule of B. The mixture is stoichiometric where no limiting reagent is present.

(vi) 5 mol A + 2.5 mol B

Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. Because of this, only 2.5 mol of B may be combined with 2.5 mol of A. There will be 2.5 mol of a remaining. B is, thus, the limiting reagent in this reaction.

(v). 2.5 mol A + 5 mol B

Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. Because of this, only 2.5 mol of A may be combined with 2.5 mol of B. There will be 2.5 mol of a remaining. A is, thus, the limiting reagent in this reaction.

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

${{\text{N}}_{\text{2}\left( \text{g} \right)}}\text{+}{{\text{H}}_{\text{2}\left( \text{g} \right)}}\to \text{2N}{{\text{H}}_{\text{3}\left( \text{g} \right)}}$

(i). Calculate the mass of ammonia produced if $\text{2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{g}$ dinitrogen reacts with $\text{1}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{g}$of dihydrogen.

Ans: First, we have to balance the given chemical equation.

${{\text{N}}_{\text{2}\left( \text{g} \right)}}\text{+3}{{\text{H}}_{\text{2}\left( \text{g} \right)}}\to \text{2N}{{\text{H}}_{\text{3}\left( \text{g} \right)}}$

From this equation we can say, 1 mole (28g) of dinitrogen reacts with 3 mole (6g) of dihydrogen to give 2 mole (34g)of ammonia.

=$\text{2}\text{.00 }\times \text{1}{{\text{0}}^{\text{3}}}\text{ g}$of nitrogen will react with$\dfrac{\text{6 g}}{\text{28 g}}\text{ }\!\!\times\!\!\text{ 2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{ g}$ dihydrogen.

=$\text{2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{g}$ of dinitrogen will react with 428.6 g of dihydrogen.

We are given the amount of dihydrogen = $\text{1}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{ g}$

Therefore,${{\text{N}}_{\text{2}}}$is the limiting reagent.

We can say that 28 g of ${{\text{N}}_{2}}$ will produce 34 g of $\text{N}{{\text{H}}_{3}}$.

Hence, mass of ammonia produced by $\text{2000 g of }{{\text{N}}_{\text{2}}}\text{ =}\dfrac{\text{34 g}}{\text{28 g}}\text{ }\!\!\times\!\!\text{ 2000 g}$

= 2428.57 g

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

Ans: ${{\text{N}}_{2}}$ is the limiting reagent while${{\text{H}}_{\text{2}}}$ is the excess reagent. So, hydrogen will be left unreacted.

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

Ans: Mass of dihydrogen left unreacted can be calculated as

$\text{=1}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{g - 428}\text{.6g}$

= 571.4 g

25. How are 0.50 mol $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ and 0.50 M $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$different?

Ans: The molar mass of $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ is given below:

$\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ = $\text{(2 }\!\!\times\!\!\text{ 23) + 12}\text{.00 + (3 }\!\!\times\!\!\text{ 16)}$

= 106 g/mol

So, 1 mole of $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ means 106 g of $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$.

Therefore, for 0.5 mol of $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ can be calculated as:

$\text{0}\text{.5 mol of N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ = }\dfrac{\text{106 g}}{\text{1 mole}}\text{ }\!\!\times\!\!\text{ 0}\text{.5 mol N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$

$\text{=53g N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$

$\text{= 0}\text{.50 M N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{=0}\text{.50mol/L N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$

Hence,$\text{0}\text{.50 mol N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$is present in 1 L of water or $\text{53g N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$is present in 1 L of water.

26. If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapor would be produced?

Ans: Let us first write the reaction between dihydrogen and dioxygen. The reaction will be:

$\text{2}{{\text{H}}_{\text{2}\left( \text{g} \right)}}\text{+}{{\text{O}}_{\text{2}\left( \text{g} \right)}}\to \text{2}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( \text{g} \right)}}$

Dioxygen reacts with two volumes of dihydrogen to generate two volumes of water vapour.

As a result, ten volumes of dihydrogen will react with five volumes of dioxygen to generate ten volumes of water Vapours.

27. Convert the following into basic units:

Ans:To convert the given numbers into basic units we have to convert it into meters and kilograms. All the numbers are converted into meters and kilograms, and are written below:

(i) 28.7 pm

Ans: $\text{1pm=1}{{\text{0}}^{\text{-12}}}\text{m}$

$\text{28}\text{.7pm=28}\text{.7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-12}}}\text{m}$

$\text{=2}\text{.87 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-11}}}\text{m}$

(ii) 15.15 pm

Ans: $\text{1pm=1}{{\text{0}}^{\text{-12}}}\text{m}$

$\text{15}\text{.15pm=15}\text{.15 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-12}}}\text{m}$

$\text{=1}\text{.515 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-12}}}\text{m}$

(iii) 25365 mg

Ans: $\text{1mg = 1}{{\text{0}}^{\text{-3}}}\text{g}$

$\text{25365mg=2}\text{.53657 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{g}$

Since,

$\text{1g=1}{{\text{0}}^{\text{-3}}}\text{kg}$

$\text{2}\text{.5365 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{1}}}\text{g=2}\text{.5365 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-1}}}\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{kg}$

$\text{25365mg=2}\text{.5365 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{kg}$$$

28. Which one of the following will have the largest number of atoms?

Ans: As we can see that all the atoms are given in grams so, by counting the number of atoms in each element we can find the answer. These are solved step by step and written below:

(i) 1 g Au (s)

Ans: $\text{1g of Au (s)=}\dfrac{\text{1}}{\text{197}}\text{mol of Au (s)}$

$\text{=}\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}{\text{197}}\text{ atoms of Au (s)}$

$\text{=3}\text{.06 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}\text{ atoms of Au (s)}$

(ii) 1 g Na (s)

Ans: $\text{1g of Na (s)=}\dfrac{\text{1}}{\text{23}}\text{mol of Na (s)}$

$\text{=}\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}{\text{23}}\text{ atoms of Na (s)}$

$\text{=0}\text{.262 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of Na (s)}$

$\text{=26}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}\text{ atoms of Na (s)}$

(iii) 1 g Li (s)

Ans: $\text{1g of Li (s)=}\dfrac{\text{1}}{\text{7}}\text{mol of Li (s)}$

$\text{=}\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}{\text{7}}\text{ atoms of Li (s)}$

$\text{=0}\text{.86 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of Li (s)}$

$\text{=86}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}\text{ atoms of Li (s)}$

(vi) 1 g of $\text{C}{{\text{l}}_{\text{2}}}$ (g)

Ans: $\text{1g of C}{{\text{l}}_{\text{2}}}\text{ (g)=}\dfrac{\text{1}}{\text{71}}\text{mol of C}{{\text{l}}_{\text{2}}}\text{ (g)}$

Molar mass of $\text{C}{{\text{l}}_{\text{2}}}$ is 71 (2 $\times $ 35.5)

\[\text{=}\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}{\text{71}}\text{ atoms of C}{{\text{l}}_{\text{2}}}\text{ (g)}\]

\[\text{=0}\text{.0848 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of C}{{\text{l}}_{\text{2}}}\text{ (g)}\]

\[\text{=8}\text{.48 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}\text{ atoms of C}{{\text{l}}_{\text{2}}}\text{ (g)}\]

Hence, 1 g of Li (s) has the largest number of atoms.

29. Calculate the molarity of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Ans: First we have to find the mole fraction of ethanol.

Mole fraction of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH=}\dfrac{\text{Number of moles }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}{\text{Number of moles of solution}}$

$\text{0}\text{.40=}\dfrac{{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}}{{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{+}{{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}}\text{ }.................\text{(1)}$

Number of moles present in 1 L water:

${{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{=}\dfrac{\text{1000 g}}{\text{18 g mo}{{\text{l}}^{\text{-1}}}}$

${{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{=55}\text{.55mol}$

Substituting the value of ${{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}$in equation (1),

$\dfrac{{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}}{{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{+55}\text{.55}}\text{=0}\text{.040}$

${{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{=0}\text{.040}{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{+}\left( \text{0}\text{.040} \right)\left( \text{55}\text{.55} \right)$

$\text{0}\text{.96}{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{=2}\text{.222mole}$

${{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{=2}\text{.314mole}$

$\text{Molarity of solution =}\dfrac{\text{2}\text{.314 mol}}{\text{1 L}}$

So, the molarity of the Ans is 2.314 M.

30. What will be the mass of one $^{\text{12}}\text{C}$ atom in g?

Ans: We know that 1 mole of carbon means 12 gram of carbon is there and in terms of number of atoms, there are $\text{6}\text{.023 }\times \text{ 1}{{\text{0}}^{\text{23}}}$ atoms in 1mole of carbon. This can be written as:

$1\text{ mole = 12 g = 6}\text{.023 }\times \text{ 1}{{\text{0}}^{23}}$

So, mass of one$^{\text{12}}\text{C}$ atom$\text{=}\dfrac{\text{12g}}{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}$

$\text{=1}\text{.993 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-23}}}\text{g}$

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

Ans: First we have to find the least precise number to find the significant figures.

(i) $\dfrac{\text{0}\text{.2856 }\!\!\times\!\!\text{ 298}\text{.15 }\!\!\times\!\!\text{ 0}\text{.112}}{\text{0}\text{.5785}}$

Ans: $\dfrac{\text{0}\text{.2856 }\!\!\times\!\!\text{ 298}\text{.15 }\!\!\times\!\!\text{ 0}\text{.112}}{\text{0}\text{.5785}}$

Least precise number of calculation = 0.112

Therefore, number of significant figures in the answer

Number of significant figures in the least precise number = 3

(ii) $\text{5 }\!\!\times\!\!\text{ 5}\text{.364}$

Ans: $\text{5 }\!\!\times\!\!\text{ 5}\text{.364}$

Least precise number of calculations = 5.364

Therefore, number of significant figures in the answer will be

Number of significant figures in 5.364 = 4

(iii) $\text{0}\text{.0125+0}\text{.7864+0}\text{.0215}$

Ans: $\text{0}\text{.0125+0}\text{.7864+0}\text{.0215}$

Since the least number of decimal places in each term is four, the number of significant figures in the answer will also be 4.

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

Isotope | Isotopic molar mass | Abundance |

$^{\text{36}}\text{Ar}$ | $\text{35}\text{.96755g mo}{{\text{l}}^{\text{-1}}}$ | 0.337% |

$^{\text{38}}\text{Ar}$ | $\text{37}\text{.96272 g mo}{{\text{l}}^{\text{-1}}}$ | 0.063% |

$^{\text{40}}\text{Ar}$ | $\text{39}\text{.9624 g mo}{{\text{l}}^{\text{-1}}}$ | 99.600% |

Ans: Molar mass of argon is calculated step by step.

$\text{=}\left[ \left( \text{35}\text{.96755 }\!\!\times\!\!\text{ }\dfrac{\text{0}\text{.337}}{\text{100}} \right)\text{+}\left( \text{37}\text{.96272 }\!\!\times\!\!\text{ }\dfrac{\text{0}\text{.063}}{\text{100}} \right)\text{+}\left( \text{39}\text{.9624 }\!\!\times\!\!\text{ }\dfrac{\text{90}\text{.60}}{\text{100}} \right) \right]\text{g mo}{{\text{l}}^{\text{-1}}}$

$\text{=}\left[ \text{0}\text{.121+0}\text{.024+39}\text{.802} \right]\text{g mo}{{\text{l}}^{\text{-1}}}$

So, the molar mass of argon is 39.947 g/ mol.

33. Calculate the number of atoms in each of the following:

Ans: All the options are given in different forms so, we have to convert it into a number of atoms. These are given below:

(i) 52 moles of Ar

Ans: $\text{1 mole of Ar = 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of Ar}$

Therefore,

$\text{52 mole of Ar = 52 }\!\!\times\!\!\text{ 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of Ar}$

$\text{=3}\text{.131 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{25}}}\text{ atoms of Ar}$

(ii) 52 u of He

Ans: $\text{1 atom of He = 4u of the He}$

Or this can be written as, 4 u of He = 1 atom of He

1. u of He = $\dfrac{\text{1}}{\text{4}}$ atom of He

of He = $\dfrac{\text{52}}{\text{4}}$ atom of He

= 13 atoms of He.

(iii) 52 g of He

Ans: $\text{4g of He = 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of He}$

Therefore, we can write:

$\text{52g of He = }\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ }\!\!\times\!\!\text{ 52}}{\text{4}}\text{ atoms of He}$

$\text{= 7}\text{.8286 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{24}}}\text{ atoms of He}$

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 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

Ans: 1 mole (44 g) of $\text{C}{{\text{O}}_{\text{2}}}$ will have 12 g carbon.

So, 3.38 g of $\text{C}{{\text{O}}_{\text{2}}}$ will have carbon = $\dfrac{\text{12g}}{\text{44g}}\text{ }\times \text{ 3}\text{.38g}$

= 0.9217 g

18 g of water will have 2 g of hydrogen.

So, 0.690 g of water contain hydrogen = $\dfrac{\text{2g}}{\text{18g}}\text{ }\times \text{ 0}\text{.690}$

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g

=0.9984 g

So, the percentage of Carbon in the compound = $\dfrac{0.9217}{0.9984}\text{ }\times \text{ 100}$= 92.32%

Now, percentage of Hydrogen in the compound = $\dfrac{0.0767}{0.9984}\text{ }\times \text{ 100}$= 7.68%

Moles of carbon in the compound = $\dfrac{92.32}{12}=7.69$

Moles of hydrogen in the compound = $\dfrac{7.68}{1}=7.68$

Since, we have the number of moles of both the elements, the ratio of carbon to hydrogen will be:

7.69: 7.68 = 1: 1

This is in the whole number, so the empirical formula will be CH.

(ii) Molar mass of the gas

Ans: We are given,

Weight of 10.0 L of the gas (at S.T.P) = 11.6 g

$\text{Weight of 22}\text{.4L of gas at STP = }\dfrac{\text{11}\text{.6g}}{\text{10}\text{.0L}}\text{ }\!\!\times\!\!\text{ 22}\text{.4L}$

$=25.984g$

$\approx 26g$

Hence, the molar mass of the gas is 26 g.

(iii) Molecular formula.

Ans: Empirical formula mass of CH = 12 + 1 = 13

Now, we can find the value of n. This is given below:

$\text{n = }\dfrac{\text{Molar mass of gas}}{\text{Empirical formula mass of gas}}$

$\text{=}\dfrac{\text{26g}}{\text{13g}}$

n = 2

Therefore,

$\text{Molecular formula of gas=}{{\left( \text{CH} \right)}_{\text{n}}}$

$\text{= }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$

35. Calcium carbonate reacts with aqueous $\text{HCl}$ to given$\text{CaC}{{\text{l}}_{\text{2}}}$ and $\text{C}{{\text{O}}_{\text{2}}}$ according to the reaction,

$\text{CaC}{{\text{O}}_{\text{3}\left( \text{s} \right)}}\text{+2HC}{{\text{l}}_{\left( \text{aq} \right)}}\to \text{CaC}{{\text{l}}_{\text{2}\left( \text{aq} \right)}}\text{+C}{{\text{O}}_{\text{2}\left( \text{g} \right)}}\text{+}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( \text{l} \right)}}$

What mass of $\text{CaC}{{\text{O}}_{\text{3}}}$ is required to react completely with 25 mL of 0.75 M HCl?

Ans: We are given the molarity as 0.75 M which means in 1 L of Ans there are 0.75 moles of HCl. This can be converted into grams as:

$\text{=}\left[ \left( \text{0}\text{.75 mol} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{36}\text{.5gmo}{{\text{l}}^{\text{-1}}} \right) \right]\text{HCl is present in 1L of water}$

$\text{=27}\text{.375g of HCl is present in 1L of water}$

Thus, 1000 mL of Ans contains 27.375 g of HCl.

So, the amount of HCl present in 25 mL of Ans

$\text{=}\dfrac{\text{27}\text{.375g}}{\text{1000mL}}\text{ }\!\!\times\!\!\text{ 25mL}$

= 0.6844 g

The chemical reaction given in the question is:

$\text{CaC}{{\text{O}}_{\text{3}\left( \text{s} \right)}}\text{+2HC}{{\text{l}}_{\left( \text{aq} \right)}}\to \text{CaC}{{\text{l}}_{\text{2}\left( \text{aq} \right)}}\text{+C}{{\text{O}}_{\text{2}\left( \text{g} \right)}}\text{+}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( \text{l} \right)}}$

$\text{2 mol of HCl }\left( \text{2 }\!\!\times\!\!\text{ 36}\text{.5=73g} \right)$reacts with $\text{1 mol of CaC}{{\text{O}}_{\text{3}}}\left( \text{100g} \right)$ .

So, the amount of$\text{CaC}{{\text{O}}_{\text{3}}}$ that will react with $\text{0}\text{.6844g = }\dfrac{\text{100}}{\text{71}}\text{ }\!\!\times\!\!\text{ 0}\text{.6844g}$

= 0.9639 g

36. Chlorine is prepared in the laboratory by treating manganese dioxide $\left( \text{Mn}{{\text{O}}_{\text{2}}} \right)$ with aqueous hydrochloric acid according to the reaction

$\text{4HC}{{\text{l}}_{\left( \text{aq} \right)}}\text{+Mn}{{\text{O}}_{\text{2}\left( \text{l} \right)}}\to \text{MnC}{{\text{l}}_{\text{2}\left( \text{aq} \right)}}\text{+C}{{\text{l}}_{\text{2}\left( \text{g} \right)}}$

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

Ans: 1 mol of $\text{Mn}{{\text{O}}_{\text{2}}}$ reacts with 4 mol of HCl.

The molar mass of $\text{Mn}{{\text{O}}_{\text{2}}}$is 87g and the molar mass of HCl is 36.5 g. For 4 moles of HCl, the mass will be 146 g.

So, for 5.0 g of $\text{Mn}{{\text{O}}_{\text{2}}}$ will react with:

$=\dfrac{146}{87}\text{ }\times \text{ 5 = 8}\text{.4 g}$

Therefore, 8.4 g of HCl will completely react with 5 g of $\text{Mn}{{\text{O}}_{\text{2}}}$.

## Chemistry Class 11 Chapter 1 - Quick Overview of Detailed Structure of Topics

Subtopic | Subtopics |

Introduction to Chemistry | Overview of the subject's scope and significance |

Importance of studying Chemistry | Understanding the relevance and applications of chemistry |

Laws of Chemical Combinations | Explanation of laws governing chemical combinations |

Dalton's Atomic Theory | Postulates and implications of Dalton's atomic theory |

Atomic and Molecular Masses | Definitions and calculations related to atomic and molecular masses |

Mole Concept and Molar Mass | Conceptual understanding and calculations involving moles and molar mass |

Percentage Composition | Determination of percentage composition in compounds |

Empirical and Molecular Formulas | Derivation and application of empirical and molecular formulas |

Chemical Equations and Stoichiometry | Balancing equations and stoichiometric calculations |

Limiting Reactants and Yield of Reaction | Concepts related to limiting reactants and theoretical yield |

Concentration of Solutions | Measurement and expression of solution concentration |

Basic Concepts of Volumetric Analysis | Introduction to volumetric analysis techniques |

## Chemistry Class 11 Chapter - 1 Important Formulas

Some important formulas of Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry are given below.

Mass % of an element =\[\frac{\textrm{Mass of that element in the compund}}{\textrm{Molar Mass of Compound}}\]X100

Mass per cent = \[\frac{\textrm{Mass of Solute}}{\textrm{Mass of Solution}}\times 100\]

Mole Fraction = \[\frac{\textrm{No. of mole of a particular component}}{\textrm{Total No. of moles of solution}}\]

Molarity = \[\frac{\textrm{No. of moles of solute}}{\textrm{Volume of solution in litres }}\]

Molarity = \[\frac{\textrm{No. of moles of solute}}{\textrm{Mass of solvent in kg}}\]

Moles of an element = \[\frac{\textrm{Mass of element}}{\textrm{Atomic Mass}}\]

Mass of one atom = \[\frac{\textrm{Atomic Mass}}{6.022\times 10^{23}}\]

Moles of a compound = \[\frac{\textrm{Mass of Compound}}{\textrm{Molecular Mass}}\]

Mass of one molecule = \[\frac{\textrm{Molecular Mass}}{6.022\times 10^{23}}\]

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The Vedantu’s Class 12 NCERT Solutions of Chemistry provided here in PDFs offer various benefits, including:

The answers provided here are straightforward.

It provides the Concise Notes and saves a lot of time for Revision.

To facilitate comprehension, solutions are presented in phases.

All of the questions from each chapter are answered.

For effective preparations, comprehend all of the processes outlined in the answers.

## Important Links for Chapter 1 Some Basic Concepts of Chemistry

Students can access extra study materials on Some Basic Concepts of Chemistry, These resources are available for download, offering additional support for your studies.

S. No | Important Links for Chapter 1: Some Basic Concepts of Chemistry |

1 | |

2 | Chapter 1: Some Basic Concepts of Chemistry Important Questions |

3 | Chapter 1: Some Basic Concepts of Chemistry NCERT Exemplar Solutions |

## Conclusion

Class 11 Chemistry Chapter 1 NCERT solutions Some Basic Concepts of Chemistry are authentic, categorised, understandable and straightforward, containing all the essential details. It helps you with completely revised solutions and critical points following exam specifications. It adheres to the newest syllabus to help you score better marks in exams. It is a difficult chapter to comprehend due to the immense number of new concepts. Since it is entirely theory-based, the solutions include step-by-step procedure, neat labelled diagrams, shortcuts, tips to approach the questions smartly. class 11 Chemistry Chapter 1 NCERT solutions are masterminded by the experts of Vedantu to serve as an ideal material for practice and make the learning process more convenient.

## Important Links for CBSE Class 11 Chemistry

S. No | Important Resources Quick Links |

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2 | |

3 | |

4 |

## FAQs on NCERT Solutions for Chemistry Class 11 Chapter 1 - Some Basic Concepts of Chemistry

1. Carbon generates two oxides, namely CO and CO_{2}, where masses of carbon which combine with Mass of oxygen are in the ratio 1:2. State how the law of multiple proportions is being followed.

The element carbon, combines with oxygen to form two different chemical compounds, which are, carbon dioxide and carbon monoxide. In CO_{2}, 12 parts of carbon mass combine with 32 parts of oxygen mass while in CO, 12 parts of carbon mass combine with 16 parts of oxygen mass. Therefore, the masses of oxygen mix with a definite mass of 12 parts of carbon in CO_{2} and CO are 32 and 16 respectively. These oxygen masses bear a simple ratio of 32: 16 or 2: 1 to each other and this is a classic example of the law of multiple proportions.

2. Define the law of multiple proportions. Explain it with an example. How does this law prove the existence of atoms?

Law of multiple proportions says that when two elements mix to generate two or more compounds, the masses of one of the elements which combine with a fixed mass of the other has an easy ratio with each other. E.g., carbon and oxygen combine to form two different chemical compounds- carbon dioxide and carbon monoxide. The masses of oxygen combining with a definite mass of carbon in CO and CO_{2} are 16 and 32, respectively. These masses of oxygen bear a ratio of 32: 16 or 2: 1 with one another. For example, sulphur and oxygen combine to form two compounds, namely, sulphur trioxide and sulphur dioxide.

3. What are some basic concepts of Chemistry?

Some basic concepts of chemistry NCERT solutions Chapter 1 of Class 11 Chemistry covers topics like Atomic and Molecular Mass, Concepts of Mass, Weight, Volume, and Temperature. The matter is categorised into compounds, mixtures, and elements. SI units of weight and temperature, the Law of Conservation of Mass, the Avogadro Law, the Law of Gaseous Volumes, etc. are also discussed. Some basic concepts of chemistry NCERT solutions cover all the basic concepts that are necessary to help students form a strong base for classes ahead and competitive exams.

4. What is the name of Chapter 1 of Chemistry Class 11?

The first Chapter of Class 11 Chemistry is called Some Basic Concepts of Chemistry. This chapter covers the basic chemistry concepts that can help the students in Class 12 and competitive exams like NEET, JEE, etc. It gives an introduction to the branches of chemistry, SI units, properties of different substances, and formulas and calculations. The students need to pay attention to this chapter to help them understand future concepts better.

5. Give an overview of questions present in NCERT Solutions for Class 11 Chemistry Chapter 1.

Some basic concepts of chemistry NCERT solutions Questions asked are short or long answer-type. Questions like calculating the atomic mass of a given element are asked, and formulas and questions asking to determine the amount of an element produced from an equation or the presence of a substance are also given. The PDF has the solutions to all the kinds of questions mentioned above and extra questions explained in an easy step-by-step manner for the students to understand the concept better.

6. Why should we download some basic concepts of chemistry class 11 NCERT solutions?

You should download the NCERT Solutions Class 11 Chemistry PDF to help you understand how to answer the questions at the end of each chapter and also refer to them if you have any confusion. Some basic concepts of chemistry class 11 NCERT Solutions PDF can also be downloaded free of cost to study offline. All answers are written and formed by expert teachers and are explained in easy language for the student to understand. The PDF also helps you familiarise yourself with other questions from within the chapter and how you should answer them in the exam. This helps you assess your knowledge of the concepts studied. These are available free of cost on Vedantu’s website and mobile app.

7. Why is class 11 Chemistry Chapter 1 solutions important?

Class 11 Chemistry Chapter 1 Solutions introduces students of the medical or non-medical stream to basic concepts of Chemistry. Concepts like atomic mass, weight, molecular mass, temperature, etc are introduced. Chemical equations and formulas based on these topics are also discussed and introduced. The students have to be thorough with these topics as they build the base and foundation for topics discussed further in the class and in the 12th. They are also important for board exams and a lot of competitive exams.

8. What are the important topics in NCERT Solutions class 11 Chemistry chapter 1?

NCERT Solutions class 11 chemistry chapter 1 covers fundamental topics such as the mole concept, atomic and molecular masses, stoichiometry, chemical equations, and the concept of equivalents. These concepts serve as the building blocks for understanding more advanced topics in chemistry.

9. What key concepts are covered in NCERT class 11 chemistry chapter 1 PDF?

NCERT class 11 chemistry chapter 1 pdf delves into fundamental topics like the mole concept, atomic and molecular masses, stoichiometry, chemical equations, and equivalents, laying the groundwork for further studies. Students can have access to other resources from chemistry chapter 1, class 11.

10. Where can I find solutions for Class 11 Chemistry Chapter 1 exercise solutions?

You can access comprehensive solutions for exercises in Chapter 1 through NCERT Solutions for Chapter 1 Chemistry class 11 or by downloading the PDF for Class 11 Chemistry Chapter 1 exercise solutions.