XI introductory chapter notes and numerical

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1. The branch of
chemistry which
deals with the
study of analysis
of obtaining,
processing,
characterizing the
composition and
structure of matter
both quantitatively
and qualitatively is
called
a. Physical
chemistry c. inorganic chemistry
b. organic chemistry d. analytical chemistry
2. What is the percent water in CuSO4.5H2O?
a. 12 b. 14
c. 16 d. 18
3. What is a mole?
a. A mole is found in a certain number of cm3 of one substance or another.
b. A mole is the sum of atomic weights.
c. A mole is the number of molecules in one gram of a substance.
d. None of the mentioned
4. Relative molecular masses can be calculated
a. By the relative atomic mass b. By the relative mass of a covalent compound
c. if individual relative atomic masses are added d. By adding relative compound masses
5. Relative molecular mass of CO2 is _________
a. 22 b. 44
c. 66 d. 88
6. Mole can be defined as atomic mass, molecular mass, formula, or ionic mass of the substance
expressed in__________.
a. Kilograms c. milligrams
b. micrograms d. grams
7. 1 mole is equal to
a. 6.02*10 23 c. 6022*1023
b. 6.02*10 -23 d. -6.022*1023
8. Excess reagent is the one which is____________, after the reaction is completed.
a. Left, unreacted c. consumed
b. Left, reacted d. none of theses

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mole
mass
solids/liquids
formula for mole calculation
n= mass/ molar.mass
n=m/mol.m
m=n*mol.mass
particles
n=no pf particlas/avagadros
numb
no of particles=n*avagadros
numb
gases
n=vol(dm3
)/22.4 dm3
.mol-1
n=vol(dm3) at stp/molar vol
Numericals
Practice problem 1.1
1. Calculate the molecular masses (in amu) of the following compounds:
(a) SO2
(b) Caffeine (C8H10N4O2)
Solution:
(a) SO2
32+2(16) =64 amu
(b) Caffeine (C8H10N4O2)
8(12) +10(1) +4(14) +2(16) = 194 amu.
2. Calculate the number of moles in 5.68 g of iron. (Atomic mass of iron Fe=55.8)
Solution:
No of moles=n=? Mass of Fe=5.68 g
As we know that n= mass/ molar. Mass so by putting values in the formula
n=5.68 g /55.8 g.mol -1 , n= 0.1 mol.
Self-assessment
1. What is the number of moles in each of the following?
(a) 52g of silicon (atomic mass of Si = 28)
Solution: As we know that n= mass/ molar. Mass so by putting values in the formula
n=52g /28 g.mol-1 n= 1.85 mol.
(b) 1.42 g of O2 (oxygen gas) (atomic mass of O= 16)
n= mass/ molar by putting values
n= 1.42 g/ 32 g/mol-1, n=0.044 mol.
3rd no) Mcqs numerical
What is the mass (in grams) of 5 moles of water (H2O)?
Solution:
No of moles of water= n= 5
Atomic mass of C=12, s=32, N=14, H=1, O=16
H2O molar mass = 2+16=18 g.mol-1

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Mass=?
So we have a very basic formula i.e. n= mass/ molar. Mass
By rearrangement of this formula we will get m=n* mol. mass
By putting values, we get m= 5*18=90 g
Exercise numerical questions
1.The mass of 5 moles of an element X is 60g.calculate the molar mass of the element. Also, name the
element.
Solution:
No of moles of X=n= 5 moles
Mass of X=m=60 g
Molar mass of X?
As we know that. n= mass/ molar. Mass
By rearrangement of this formula we will get mol. Mass= m/n
By putting values mol. mass= 60/5=12 g.mol-1
The element X is carbon atom. See periodic table or text book page no 4 for reference
2.Calculate the molecular or formula masses of the following compounds.
(a) C2H5OH (b) Al2O3 (c) K2Cr2O7
Solution:
(a) C2H5OH
2(12) +5(1) +16+1=46
amu
(b) Al2O3
2(27) +3(16) = 102 amu
(c) K2Cr2O7
2(39) +2(52) +7(16)
=294 amu
3. Calculate the mass in grams of (a) 7.75 moles of Al2O3 (b) 15 moles of H2SO4
Solution:
(a) 7.75 moles of Al2O3
As we know that n= mass/ molar. Mass
n= 7.75 moles, molar mass of Al2O3 = 2(27) +3(16) = 102 g/mol
after rearrangement and then putting values in formula
m=n*mol. mass so m= 7.75* 102= 790.5 g
(b) 15 moles of H2SO4
molar mass of H2SO4 = 2(1) +32+4(16) = 98 g/mol
n= mass/ molar. Mass, after rearrangement and then putting values in formula
Atomic mass pf C= 12, K=39, Cr=52, Al= 27, H=1, O=16
Atomic masses of S= 32, Al=27, H=1, O=16

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m=n*mol. Mass= 15* 98=1470 g.
4. how many moles are present in each of the following samples?
(a)
30 g of
MgS
Solution: n= mass/ molar. Mass, by putting values
Molar mass of MgS= 24+32=56 g/mol
n=30/56= 0.535 mol
(b) 75 g of Ca
n= mass/ molar. Mass= 75/40= 1.875 mol.
(c) 8.85 kg of CO2
In mole, mass is reported in grams, here above value is in kgs, so move for grams instead of kgs.
1kg = 1000 g
8.85kg = ?
1kg*? = 1000 * 8.85
? = 1000 * 8.85/1 = 8850 g
Molar mass of CO2= 12+2(16) = 44 g/mol
n= mass/ molar. Mass, n=8850 / 44 = 201.13 mol
(d) 40 g of NaCl
Molar mass of NaCl = 23+ 35.5 = 58.5 g/mol, n= mass/ molar. Mass
n=40 / 58.5 = 0.6837 mol
********** End of moles for solids/liquids***********
Numericals
Practice problem 1.1
How many Cu atoms are present in 0.5 mol of copper?
Solution:
As we know that n= no of particles/ Avogadro’s number
Rearrange and put values in above formula
No of particles = n* Avogadro’s number
No of particles=0.5* 6.02I*1023 = 3.01*1023
Practice problem 1.2
What is the mass of 1.204*1022 atoms of lead? (atomic mass of Pb=207)
Solution:
Atomic masses of S= 32, C=12, H=1, O=16, Na= 23, Mg= 24, Cl= 35.5 , Ca= 40
Mole
particles

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No of particles=1.204*1022 atoms of lead Mass of lead=?
As n= mass/ molar. Mass & n= no of particles/ Avogadro’s number
To calculate mass, find out n by using above formula then put values
so by putting values in the formula n= no of particles/ Avogadro’s number
n= 1.204*1022 atoms of lead/6.02*1023=0.0169 mol
n=mass/ molar. Mass or mass= n*molar mass= 0.016*207= 3.498g
Self-assessment
1. What is the number of moles in each of the following?
(c) 3.6*1024 atoms of lithium (atomic mass of li = 7)
Solution:
n= no of particles/ Avogadro’s number = 3.6*1024 atoms of lithium/6.02*1023
n=5.98 mol
(d) 4*1022 formula units of KCl. K=39, Cl=35.5
n= no of particles/ Avogadro’s number=4*1022/6.02*1023=0.066 mol
2. Which of the following contains greatest number of particles?
(a) 4g of lithium atomic mass=7 n= mass/ molar.
Mass
n= no of particles/
Avogadro’s
number
n=4/7=0.5714
no of particles=n*NA
no of particles=0.5714*6.02*1023
no of particles=3.44*1023
(b) 4g of Cl2 atomic mass=35.5 n= mass/ molar.
Mass
n= no of particles/
Avogadro’s
number
n=4/71=0.0563
no of particles=n*NA
no of particles=0.0563*6.02*1023
no of particles=0.399*1023
(c) 4g of H2 atomic mass=1 n= mass/ molar.
Mass
n= no of particles/
Avogadro’s
number
n=4/1=4
no of particles=n*NA
no of particles=4*6.02*1023
no of particles=24.08*1023
(d) 4g of H2O atomic mass=18 n= mass/ molar.
Mass
n= no of particles/
Avogadro’s
number
n=4/18=0.222
no of particles=n*NA
no of particles=0.222*6.02*1023
no of particles=1.337*1023
4th no) Mcqs numerical
The number of molecules in 22g of CO2?
Solution:
n= no of particles/ Avogadro’s number & n= mass/ molar. Mass
n=22 g /44 g.mo-1= 0.5 mol
no of particles=n*NA= 0.5*6.02*1023 = 3.011*1023
Exercise numerical questions
3c. Calculate the mass in grams of 1.0*1025 molecules of H2O.
Solution:

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n= mass/ molar. Mass & n= no of particles/ Avogadro’s number
to find no of moles use n= no of particles/ Avogadro’s number
by putting values
n= 1.0*1025 molecules of H2O/6.02*1023
n=16.60 mol
rearrange n= mass/ molar. Mass
m=n*molar. mass, m= 16.60*18= 298.8 g
4e. How many moles are present in each of the following?
(e) 7.5*1020 molecules of C6H6
Solution:
n= no of particles/ Avogadro’s number, by putting values
n=7.5*1020 molecules of C6H6 /6.02*1023
n=1.245*10-3 mol.
********** End of moles for particles***********