NCERT Solutions Class 11 Chemistry Chapter 1 Some Basic Concepts Of Chemistry – Here are all the NCERT solutions for Class 11 Chemistry Chapter 1. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Some Basic Concepts Of Chemistry of Chemistry taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Chemistry, then you must come across chapter 1 Some Basic Concepts Of Chemistry. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts Of Chemistry in one place.
NCERT Solutions Class 11 Chemistry Chapter 1 Some Basic Concepts Of Chemistry
Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Chemistry for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.
For a better understanding of this chapter, you should also see summary of Chapter 1 Some Basic Concepts Of Chemistry , Chemistry, Class 11.
Class | 11 |
Subject | Chemistry |
Book | Chemistry Part I |
Chapter Number | 1 |
Chapter Name |
Some Basic Concepts Of Chemistry |
NCERT Solutions Class 11 Chemistry chapter 1 Some Basic Concepts Of Chemistry
Class 11, Chemistry chapter 1, Some Basic Concepts Of Chemistry solutions are given below in PDF format. You can view them online or download PDF file for future use.
Some Basic Concepts Of Chemistry Download
Did you find NCERT Solutions Class 11 Chemistry chapter 1 Some Basic Concepts Of Chemistry helpful? If yes, please comment below. Also please like, and share it with your friends!
NCERT Solutions Class 11 Chemistry chapter 1 Some Basic Concepts Of Chemistry- Video
You can also watch the video solutions of NCERT Class11 Chemistry chapter 1 Some Basic Concepts Of Chemistry here.
Video – will be available soon.
If you liked the video, please subscribe to our YouTube channel so that you can get more such interesting and useful study resources.
Download NCERT Solutions Class 11 Chemistry chapter 1 Some Basic Concepts Of Chemistry In PDF Format
You can also download here the NCERT Solutions Class 11 Chemistry chapter 1 Some Basic Concepts Of Chemistry in PDF format.
Click Here to download NCERT Solutions for Class 11 Chemistry chapter 1 Some Basic Concepts Of Chemistry
Question & Answer
Q.1: Calculate the molar mass of the following:
(i) \( \mathrm{H}_{2} \mathrm{O} \)
(ii) \( \mathrm{CO}_{2} \)
(iii) \( \mathrm{CH}_{4} \)
Ans : (i) \( \mathrm{H}_{2} \mathrm{O} \): The molecular mass of water, \( \mathrm{H}_{2} \mathrm{O} \) \( =(2 \times \text { Atomic mass of hydrogen) }+(1 \times \text { Atomic mass of oxygen }) \) \( =[2(1.0084)+1(16.00 \mathrm{u})] \) \( =2.016 \mathrm{u}+16.00 \mathrm{u} \) \( \begin{aligned} &=18.016 \\ &=18.02 \mathrm{u} \end{aligned} \) (ii) \( \mathrm{CO}_{2} \) The molecular mass of carbon dioxide, \( \mathrm{CO}_{2} \) \( \begin{aligned} &=(1 \times \text { Atomic mass of carbon })+(2 \times \text { Atomic mass of oxygen }) \\ &=[1(12.011 \mathrm{u})+2(16.00 \mathrm{u})] \\ &=12.011 \mathrm{u}+32.00 \mathrm{u} \\ &=44.01 \mathrm{u} \end{aligned} \) (iii) \( \mathrm{CH}_{4 :} \) The molecular mass of methane, \( \mathrm{CH}_{4} \) \( \begin{array}{l}{=(1 \times \text { Atomic mass of carbon })+(4 \times \text { Atomic mass of hydrogen })} \\ {=[1(12.011 \mathrm{u})+4(1.008 \mathrm{u})]} \\ {=12.011 \mathrm{u}+4.032 \mathrm{u}} \\ {=16.043 \mathrm{u}}\end{array} \)
Q.2: Calculate the mass percent of different elements present in sodium sulphate \( \left(\mathrm{Na}_{2} \mathrm{SO}_{4}\right) \).
Ans : The molecular formula of sodium sulphate is \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) Molar mass of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) \( =[(2 \times 23.0)+(32.066)+4(16.00)] \) \( =142.066 \mathrm{g} \) \( \text{Mass percent of an element} \) \( =\frac{\text { Mass of that element in the compound }}{\text { Molar mass of the compound }} \times 100 \) \( \therefore \) Mass percent of sodium: \( =\frac{46.0 \mathrm{g}}{142.066 \mathrm{g}} \times 100 \) \( \begin{array}{l}{=32.379} \\ {=32.4 \%}\end{array} \) Mass percent of sulphur: \( =\frac{32.066 \mathrm{g}}{142.066 \mathrm{g}} \times 100 \) \( \begin{aligned} &=22.57 \\ &=22.6 \% \end{aligned} \) Mass percent of oxygen: \( =\frac{64.0 \mathrm{g}}{142.066 \mathrm{g}} \times 100 \) \( \begin{array}{l}{=45.049} \\ {=45.05 \%}\end{array} \)
Q.3: Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.
Ans : \( \% \text { of iron by mass }=69.9 \%[\text { Given }] \) \( \% \text { of oxygen by mass }=30.1 \%[\text { Given }] \) \( \text{Relative moles of iron in iron oxide:} \) \( =\frac{\% \text { of iron by mass }}{\text { Atomic mass of iron }} \) \( =\frac{69.9}{55.85} \) \( \begin{array}{l}{=1.25} \\ {\text { Relative moles of oxygen in iron oxide: }}\end{array} \) \( =\frac{\% \text { of oxygen by mass }}{\text { Atomic mass of oxygen }} \) \( =\frac{30.1}{16.00} \) \( =1.88 \) Simplest molar ratio of iron to oxygen: \( \begin{array}{l}{=1.25 : 1.88} \\ {=1 : 1.5} \\ {=2 : 3}\end{array} \) \( \therefore \text { The empirical formula of the iron oxide is } \mathrm{Fe}_{2} \mathrm{O}_{3} \)
Q.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.
Ans : The balanced reaction of combustion of carbon can be written as: (i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to Produced 1 mole of carbon dioxide. (ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant. (iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.
Q.5: Calculate the mass of sodium acetate \( \left(\mathrm{CH}_{3} \text { COONa) }\right. \) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 \( \mathrm{g} \mathrm{mol}^{-1} \).
Ans : \( 0.375 \mathrm{M} \) aqueous solution of sodium acetate. \( \equiv 1000 \mathrm{mL} \text { of solution containing } 0.375 \) moles of sodium acetate. \( \therefore \) Number of moles of sodium acetate in 500 mL. \( =\frac{0.375}{1000} \times 500 \) \( =0.1875 \) mole Molar mass of sodium acetate = \( 82.0245 \mathrm{g} \mathrm{mole}^{-1}(\text { Given }) \) \( \therefore \text { Required mass of sodium acetate }=\left(82.0245 \mathrm{g} \mathrm{mol}^{-1}\right)(0.1875 \text { mole) } \) \( =15.38 \mathrm{g} \)
NCERT / CBSE Book for Class 11 Chemistry
You can download the NCERT Book for Class 11 Chemistry in PDF format for free. Otherwise you can also buy it easily online.
- Click here for NCERT Book for Class 11 Chemistry
- Click here to buy NCERT Book for Class 11 Chemistry
All NCERT Solutions Class 11
- NCERT Solutions for Class 11 Accountancy
- NCERT Solutions for Class 11 Biology
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Economics
- NCERT Solutions for Class 11 History
- NCERT Solutions for Class 11 Geography
- NCERT Solutions for Class 11 Political Science
- NCERT Solutions for Class 11 Sociology
- NCERT Solutions for Class 11 Psychology
- NCERT Solutions for Class 11 English
- NCERT Solutions for Class 11 Hindi
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Business Studies
- NCERT Solutions for Class 11 Statistics
All NCERT Solutions
You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Class 4 | Class 5 | Class 6 |
Class 7 | Class 8 | Class 9 |
Class 10 | Class 11 | Class 12 |
Download the NCERT Solutions app for quick access to NCERT Solutions Class 11 Chemistry Chapter 1 Some Basic Concepts Of Chemistry. It will help you stay updated with relevant study material to help you top your class!
To get fastest exam alerts and government job alerts in India, join our Telegram channel.
Discussion about this post