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

by aglasem
June 20, 2019
in 11th Class
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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.

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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

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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

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