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NCERT Solutions Class 11 Chemistry Chapter 2 Structure Of Atom – Here are all the NCERT solutions for Class 11 Chemistry Chapter 2. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Structure Of Atom 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 2 Structure Of Atom. After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Chemistry Chapter 2 Structure Of Atom in one place.

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NCERT Solutions Class 11 Chemistry Chapter 2 Structure Of Atom

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Class 11
Subject Chemistry
Book Chemistry Part I
Chapter Number 2
Chapter Name

Structure Of Atom

NCERT Solutions Class 11 Chemistry chapter 2 Structure Of Atom

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Question & Answer

Q.1: (i) Calculate the number of electrons which will together weigh one gram. 
(ii) Calculate the mass and charge of one mole of electrons.

Ans : (i) Mass of one electron = \( 9.10939 \times 10^{-31} \mathrm{kg} \)
\( \therefore \text { Number of electrons that weigh } 9.10939 \times 10^{-31} \mathrm{kg}=1 \)
Number of electrons that will high \( 1 \mathrm{g}=\left(1 \times 10^{-3} \mathrm{kg}\right) \) 
\( =\frac{1}{9.10939 \times 10^{-31} \mathrm{kg}} \times\left(1 \times 10^{-3} \mathrm{kg}\right) \)
\( =0.1098 \times 10^{-3+31} \)
\( \begin{aligned} &=0.1098 \times 10^{28} \\ &=1.098 \times 10^{27} \end{aligned} \)
(ii) Mass of one electron = \( 9.10939 \times 10^{-31} \mathrm{kg} \)
Mass of one mole of electron = \( 9.10939 \times 10^{-31} \mathrm{kg} \)
Mass of one mole of electron = \( \left(6.022 \times 10^{23}\right) \times\left(9.10939 \times 10^{-31} \mathrm{kg}\right) \)
\( =5.48 \times 10^{-7} \mathrm{kg} \)
Charge on one electron \( =1.6022 \times 10^{-19} \) coulomb
Charge on one mole of electron = \( \left(1.6022 \times 10^{-19} \mathrm{C}\right)\left(6.022 \times 10^{23}\right) \) 
\( =9.65 \times 10^{4} \mathrm{C} \) 
Q.2: (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of \( ^{14} \mathrm{C} \).
( (Assume that mass of a neutron = \( 1.675 \times 10^{-27} \mathrm{kg} \) ).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of \( \mathrm{NH}_{3} \) at STP.
Will the answer change if the temperature and pressure are changed ?

Ans : (i) Number of electrons present in 1 molecule of methane \( \left(\mathrm{CH}_{4}\right) \) 
\( \{1(6)+4(1)\}=10 \)
Number of electrons present in 1 mole i.e., \( 6.023 \times 10^{23} \) molecules of methane \( =6.022 \times 10^{23} \times 10=6.022 \times 10^{24} \) 


(ii) (a) NUmber of atoms of \( ^{14} \mathrm{C} \text { in } 1 \) mole = \( 6.023 \times 10^{23} \) 
\( \begin{array}{l}{\text { since } 1 \text { atom of }^{14} \mathrm{C} \text { contains }(14-6) \text { i.e., } 8 \text { neutrons, the number of neutrons in } 14 \mathrm{g} \text { of }} \\ {^{14} \mathrm{C} \text { is }\left(6.023 \times 10^{23}\right) \times 8 . \text { Or, } 14 \mathrm{g} \text { of }^{14} \mathrm{C} \text { contains }\left(6.022 \times 10^{23} \times 8\right) \text { neutrons. }}\end{array} \)
\( \begin{array}{l}{\text { Number of neutrons in } 7 \mathrm{mg}} \\ {=\frac{6.022 \times 10^{23} \times 8 \times 7 \mathrm{mg}}{1400 \mathrm{mg}}}\end{array} \)
\( =2.4092 \times 10^{21} \) 
(b) Mass of one neutron = \( 1.67493 \times 10^{-27} \mathrm{kg} \) 
Mass of total neutron in \( 7 \mathrm{g} \text { of }^{14} \mathrm{C} \) 
\( \begin{array}{l}{=\left(2.4092 \times 10^{21}\right)\left(1.67493 \times 10^{-27} \mathrm{kg}\right)} \\ {=4.0352 \times 10^{-6} \mathrm{kg}}\end{array} \)


(iii) (a) 1 mole of \( \mathrm{NH}_{3}=\{1(14)+3(1)\} \mathrm{g} \text { of } \mathrm{NH}_{3} \)
\( \begin{array}{l}{=17 \mathrm{g} \text { of } \mathrm{NH}_{3}} \\ {=6.022 \times 10^{23} \mathrm{molecules} \text { of } \mathrm{NH}_{3}}\end{array} \)
Total number of protons present in 1 molecule of \( \mathrm{NH}_{3} \) 
\( \begin{array}{l}{=\left(6.023 \times 10^{23}\right)(10)} \\ {=6.023 \times 10^{24}}\end{array} \)
Number of protons in \( 6.023 \times 10^{23} \) molecules of \( \mathrm{NH}_{3} \) 
\( \begin{array}{l}{=\left(6.023 \times 10^{23}\right)(10)} \\ {=6.023 \times 10^{24}}\end{array} \) 
\( \Rightarrow 17 \text { g of } \mathrm{NH}_{3} \text { contains }\left(6.023 \times 10^{24}\right) \) protons 
Number of protons in 34 mg of \( \mathrm{NH}^{3} \) 
\( =\frac{6.022 \times 10^{24} \times 34 \mathrm{mg}}{17000 \mathrm{mg}} \) 
\( =1.2046 \times 10^{22} \)
(b) Mass of one proton = \( 1.67493 \times 10^{-27} \mathrm{kg} \)
Total mass of protons in 34 mg of \( \mathrm{NH}_{3} \)
\( \begin{array}{l}{=\left(1.67493 \times 10^{-27} \mathrm{kg}\right)\left(1.2046 \times 10^{22}\right)} \\ {=2.0176 \times 10^{-5} \mathrm{kg}}\end{array} \) 
The number of protons, electrons, and neutrons in an atom is independent of 
temperature and pressure conditions. Hence, the obtained values will remain unchanged 
if the temperature and pressure is changed. 
Q.3: How many neutrons and protons are there in the following nuclei ?
\( _{6}^{13} C \), \( _{8}^{16} \mathrm{O} \), \( _{12}^{24} \mathrm{Mg} \), \( \begin{array}{l}{56} \\ {26}\end{array} \mathrm{F} \mathrm{e} \), \( \underset{38}{88} \mathrm{Sr} \).

Ans : \( _{6}^{13} \mathrm{C} \):
Atomic mass = 13 
Atomic number = Number of protons = 6 
Number of neutrons = ( Atomic mass ) - (Atomic number)
\( =13-6=7 \) 
 \( _{8}^{16} \mathrm{O} \):
Atomic mass = 16
Atomic number = 8 
Number of protons = 8 
Number of neutrons = (Atomic mass)-(Atomic number)
\( =16-8=8 \)
\( _{12}^{24} \mathrm{Mg} \) : 
Atomic mass = 24 
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) - (Atomic number)
\( =24-12=12 \)
\( \begin{array}{l}{56} \\ {26}\end{array} \mathrm{F} \mathrm{e} \):
\( \begin{array}{l}{\text { Atomic mass }=56} \\ {\text { Atomic number }=\text { Number of protons }=26} \\ {\text { Number of neutrons }=(\text { Atomic mass })-(\text { Atomic number })} \\ {=56-26=30}\end{array} \) 
\( \underset{38}{88} \mathrm{Sr} \): 
\( \begin{array}{l}{\text { Atomic mass }=88} \\ {\text { Atomic number }=\text { Number of protons }=38} \\ {\text { Number of neutrons = (Atomic mass)-(Atomic number) }} \\ {=88-38=50}\end{array} \) 
Q.4: Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A).
(i) Z = 17 , A = 35. 
(ii) Z = 92 , A = 233. 
(iii) Z = 4 , A = 9.

Ans : (i) \( _{17}^{35} \mathrm{Cl} \)
(ii) \(  _{92}^{233} \mathrm{U} \)
(iii) \( _{4}^{9} \mathrm{Be} \) 
Q.5: Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ν ) of the yellow light.

Ans : From the expression,
\( \lambda=\frac{c}{v} \)
We get,
\( v=\frac{c}{\lambda}_{\ldots \ldots \ldots(1)} \)
Where,
\( v= \) frequency of yellow light
\( c=\text { velocity of light in vacuum }=3 \times 10^{8} \mathrm{m} / \mathrm{s} \) 
\( \lambda=\text { wavelength of yellow light }=580 \mathrm{nm}=580 \times 10^{-9} \mathrm{m} \)
Substituting the values in expression(i):
\( v=\frac{3 \times 10^{8}}{580 \times 10^{-9}}=5.17 \times 10^{14} \mathrm{s}^{-1} \)
Thus, frequency of yellow light emitted from the sodium lamp
\( =5.17 \times 10^{14} \mathrm{s}^{-1} \)
Wave number of yellow light, \( \overline{v}=\frac{1}{\lambda} \) 
\( =\frac{1}{580 \times 10^{-9}}=1.72 \times 10^{6} \mathrm{m}^{-1} \)

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