Ppt basic concepts in chemistry, xi, Dr Mona Srivastava Founder - MasterChem Classes

CLASS-11
SUBJECT- CHEMISTRY
CHAPTER - Basic Concepts in
Chemistry
Dr. Mona Srivastava
M.Sc. Ph. D. Chemistry

*Period 1
Introduction to the topic.
Importance and role of chemistry in our lives.
Some basic calculations and laws of chemistry.
Learning objectives

The branch of science concerned with the
substances of which matter is composed,
the investigation of their properties and
reactions, and the use of such reactions
to form new substances.

Chemistry related to different branches

Assignment no. 1
Q.1. Fill in the blanks with following conversions-
(i) 1km = -------mm = -------pm
(ii) 1mg = -------kg = -------ng
(iii) 1 ml = ------L = -------m3 ( P 27, Q.1.21 b ncert)
(Hint – Refer to table no .1.3 , p-9, p-10 ncert )
Answer-
(i) 1km = 10 6 mm , = 10 15 pm
(ii) 1mg = 10 -6 kg = 10 6 ng
(iii) 1ml = 10 -3 L = 10 -3 dm 3

*Period -2
*Learning objectives-
*Scientific notations

 Chemists often work with numbers that are extremely large or extremely
small.
 For example, there are 10,300,000,000,000,000,000,000 carbon atoms in
a 1-carat diamond each of which has a mass of
0.000,000,000,000,000,000,000,020 grams.
 It is impossible to multiply these numbers with most calculators because
they can't accept either number as it is written here.
 To do a calculation like this, it is necessary to express these numbers in
scientific notation, as a number between 1 and 10 multiplied by 10 raised to
some exponent.

Scientific Notation
Scientific notation (also referred to as scientific form
or standard index form, or standard form in the UK)
is a way of expressing numbers that are too big or
too small to be conveniently written in decimal form.

How to add and subtract numbers in scientific notation with different
exponents ?
Step 1 : Adjust the exponents of 10 in the given numbers such that
they have the same exponent.
(Tip : Always it is easier to adjust the smaller exponent to equal the
larger exponent).
Step 2 : In step 2, you will have the same exponent for 10 in all the
numbers. For example, 10n So, factor 10n out from all the numbers.
Step 3 : Now, add or subtract the numbers and write the final answer
in scientific notation.

Adding and Subtracting with Scientific Notation
How to add and subtract numbers in scientific notation with same
exponent ?
Let us consider the addition of two numbers in scientific notation.
(9.35 x 103) + (8.34 x 103)
First take out the factor of 103 common .
Then, value becomes = (8.35 + 9.34) x 103 = 17.69 x 103
Now write the above number in scientific notation = 1.769 x 104
Therefore, (9.35 x 103) + (8.34 x 103) = 17.69 x 103
Note : The same process has to be followed for subtracting
numbers in scientific notation.

Multiplying and dividing with scientific notation :
We can multiply and divide numbers with scientific notation by using
properties of exponents as follows –
a.Multiplication of scientific notations-
Example 1 :
Multiply : (3.2 x 105) x (2.67 x 103)
Solution :
= (3.2 x 2.67) x (105 x 103) = (8.544) x (105+3) = 8.544 x 108
The above number is in scientific notation. Therefore,
(3.2 x 105) x (2.67 x 103) = 8.544 x 108

b. Division of scientific notations-
Example - (2.688 x 106) / (1.2 x 102)
Give your answer in scientific notation.
Solution : = (2.688 / 1.2) x (106 / 102)
= (2.24) x (106-2)
= 2.24 x 104
Therefore, (2.688 x 106) / (1.2 x 102) = 2.24 x 104

*Q.1. Express the following in scientific notations-
(i) 0.0048 ,(ii) 234.000 , (iii) 8008 , (iV) 500.0 (V)6.0012.
Q.2. Express the following calculations in scientific notations after
solving them.
1. Addition – ( 665 x104 ) + (0.895 x 103) ( ans 3.864 x 1014 )
2. Subtraction – (2.5x 10-2) - ( 4.8 x 10-3) ( ans 2.02 x 10 -2
)
3. Multiplication -(5.6 x 105 ) x (6.9 x 10 8 ) ( ans 3.864 x 1014 )
4. Division – 2.7 x 10
-3
/ 5.5 x 10
4
( ans 4.909 x 10
-8
)
Assignment no.2

*PERIOD 3
*Learning Objective –
1.Significant figures
2.Precision
3.Accuracy

*What are accuracy , accuracy and
significant figures ?
*Accuracy “ refers to how closely a measured
value agrees with the correct value. “
*Precision “refers to how closely individual
measurements agree with each other. ...”
*significant figures “ are the number of digits
believed to be correct by the person doing the
measuring.”

 Significant figures are those meaningful digits that are known with
certainty.
 They indicate uncertainty in an experiment or calculated value.
 For example if 15.6 ml is the result of an experiment, then 15 is
certain while 6 is uncertain, and the total number of significant
figures are 3.
 Significant figures are defined as the total number of digits in a
number including the last digit that represents the uncertainty of
the result.

Significant digits are certain digits that have significance or
meaning and give more precise details about the value of the
number.

We must remember about significant figures that ----
-----

5 RULES FOR SIGNIFICANT FIGURES—
1. All non-zero numbers ARE significant. ...
2. Zeros between two non-zero digits ARE significant.
3. Leading zeros are NOT significant. ...
4. Trailing zeros to the right of the decimal are
significant. ...
5. Trailing zeros in a whole number with the decimal
shown ARE significant.

* Example – Lets understand by following example.
*Rules for calculation with significant figures

Q. Add the following numbers and report your answer in correct significant figures.
a. 39.64 +1.3 b.195.4-193 c. d.46.8-41. d. 900 +500
Answer-
a. Total is 39.64 +1.3 = 40.94 but the limiting factor is 1.3 therefore we will
repot with only one decimal
Answer- 40.9
b. The value after subtraction is 2.4 but the limiting factor is 193 therefore
the answer = 2
c. Here both the numbers in the question only one decimal place therefore the
answer should also have only one decimal place Answer = 5.8
d. In this problem 900 and 500 both have trailing 2 zeros which are not
significant . Therefore the answer should be limited to hundreds place . The
total is 1400 ,which has 2 trailing zeros .
The answer should be expressed in scientific notation as- 1,4 x10 3
Try yourself

*Assignment no. 3
Q.1. How many significant figures are present in the following?
(i) 0.0025 , (ii) 208 , (iii) 5005 , (iv) 126,000 , (v) 500.0 , (vi)
2.0034 . (Q.1.19 ncert p26)
Ans. (i) 0.0025 There are 2 significant figures.
(ii) 208 There are 3 significant figures.
(iii) 5005 There are 4 significant figures
(iv) 126,000 There are 3 significant figures.
(v) 500.0 There are 4 significant figures.
(vi) 2.0034 There are 5 significant figures.

Q.2.Round up the following figures up to 3 significant figures.
(i) 34.216 ,(ii) 10.4107 ,(iii) 0.04597 ,(iv) 2808 (Q.1.20 ncert p27)
Ans. (i) 34.2 ,
(ii) 10.4 ,
(iii) 0.0460 ,
(iv) 2810
Q.3. How many significant figures should be present in the
answer of the following calculations? ( Q.1.31 ncert P 28)
(i) 0.02856 x 298 X 0.112
0.5785
(ii) 5 × 5.364 , (iii) 0.0125 + 0.7864 + 0.0215

Solution- (i) Least precise number of calculation = 0.112
Number of significant figures in the answer
= Number of significant figures in the least precise number = 3
(ii) 5 × 5.364
Least precise number of calculation = 5.364
Therefore, Number of significant figures in the answer = Number
of significant figures in 5.364 = 4
(iii) 0.0125 + 0.7864 + 0.0215
Since the least number of decimal places in each term is four,
the number of significant
figures in the answer is also 4

*PERIOD 4
*Learning objectives –
“Dimensional analysis ”

* What do you mean by dimensional analysis ?
*Dimensional Analysis ( the Unit Factor Method)
is a problem-solving method that uses the fact
that----
“ any number or expression can be multiplied by
one without changing its value. ”

* Example for learning Some basic conversions in dimensional analysis
*Q.1 .A man weighs 175 pounds .Calculate his weight in Kg given (1kg
= 2.205 pounds )
*Solution . Since 1kg = 2.205 pounds (lb) 1kg = 2.205 pounds
*Therefore 1 pound = 1/ 2.205 kg = 0.453 kg.
*Therefore 175 pounds = 175 x 0.453 kg = 79.4 kg.
*Q.2 . Convert the density given in g/cm3 into kg/m3
*Solution - As we know that 1kg = 1000 g and 1m = 100 cm.
*So we can write as 1 g = 0.001 kg and 1cm = 0.01 m
*Now converting the unit- 1g / cm 3 = 0.001kg /( 0.01) 3 m 3
* = 0.001 / 0.000001 = 1 / 0.0001 = 1000 kg /m3

* Solution –
For converting 1L into SI unit following steps are
followed-
The SI unit for volume is m3.
1 L = 1000 cm 3 = 1000 (cm x cm x cm )
= 1000 x 1/100 m x 1/100m x 1/100 m
= 1000/ 1000000 cm3 = 10 -3 m3
Q.3 . Convert volume of 1 L into SI unit.

Practice time
Q.1 Calculate the mass in grams of 14.79 ml of a substance. Its
density is 1.193 g/ml.
Solution:
1.) The given measurement is 14.79 ml.
2.) It must be converted to grams.
3.) The conversion factor has to relate mass and volume. Luckily,
density is supplied, and its units are g/ml, a mass and volume unit. The
number 1.193 belongs to the gram unit, and is placed on the top so that
the ml units can cancel out.
4) 14.79 ml 1.193 g 17.64 g
-------- X ---- = ----- = 17.64 g (significant figures applied)
1 ml 1

Answer the following questions?
1. How many pounds does 2.00 kg of cheese weigh?
2. How many mL are in 0.50 quarts?
3. How many inches are in 1.00 km?
4. How many feet are in 3.45 km?
5. How many km are in 5.00 miles?
6. How many yards are in 72.5 miles?
7. How many cc's are in 979 mL?
8. How many Kelvins are in -15.5oC?
9. How many degrees Celsius are in 315 K?
10. How many degrees Fahrenheit are in 30.0oC?
* ASSIGNMENT NO 4

*Answers to the assignment questions
* 1) . 2.00 kg (1000g / 1 kg)(1 lb / 454 g) = 4.41 pounds
* 2) 0.50 quarts (946 mL / quart) = 470 mL
* 3). 1.00 km (1000 m / 1 km)( 100 cm / 1 m)(1 inch / 2.54 cm) = 3.94 x 104
* or 39,400 inches
* 4) 3.45 km (1000 m / 1 km)( 100 cm / 1 m)(1 inch / 2.54 cm)(1 ft / 12 in) = 11,300
or 1.13 x 104
* feet
* 5) 5.00 mile(1760 yds / 1 mile)(3 ft / 1 yd)(12 in / 1 ft)(2.54 cm / 1 in)(1 m /
100cm)(1 km / 1000m ) = 8.05
* km
* 6) 72.5 miles (1760 yds / 1 mile ) = 1.28 x 105
* or 128,000 yards
* 7) 979 cc (cc = mL)
* 8) -15.5oC + 273 = 258K
* 9) 315K - 273 = 42oC
* 10) 10. 86o F

*Period 5
*Basic laws of chemical combination
Learning objective

Law of conservation of mass -
(Antoine Lavoisier)

*Problem for Practice
*ANS-According to law of conservation of mass,
*Total mass of reactants = Total mass of products.
*Here, baking soda mixture (reactant) on heating gives solid
residue and carbon dioxide ( products).Here baking soda mixture
(reactant) on heating gives solid residue and carbon dioxide (
products).
*M Baking soda = M Solid residue + M Carbon dioxide
*Hence, the mass of solid residue is 150g – 87g = 63g.
Q. In a chemical reaction 150 g Baking soda mixture
containing sodium bicarbonate and vinegar on heating gives
87 g of carbon dioxide gas. What mass of solid residue will
left in food?

ASSIGNMENT NO. 5 (Question 1.21: Ncert as homework no.1 )
Q. The following data are obtained when dinitrogen and dioxygen react together to
form different compounds:
Mass of dinitrogen Mass of dioxygen
(i) 14 g 16 g
(ii) 14 g 32 g
(iii) 28 g 32 g
(iv) 28 g 80 g
Which law of chemical combination is obeyed by the above experimental data?
Give its statement.

*Period 6
*Learning objective
*Gaylussacs Law of equal volume
* Avogadrovs law

*It states that when gases
combine or are produced in a
chemical reaction they do so
in a simple ratio by volume,
provided all gases are at the
same temperature and
pressure.

* Question 1.26: ( from ncert)
If ten volumes of dihydrogen gas react with five volumes of
dioxygen gas, how many volumes of water vapour would be
produced
*Numerical related to law
*Answer
*Reaction of dihydrogen with dioxygen can be written as:
2H2 + O2 = 2H2O
*Now, two volumes of dihydrogen react with one volume of
dihydrogen to produce two volumes of water vapour.
*Hence, ten volumes of dihydrogen will react with five volumes of
dioxygen to produce ten volumes of water vapour.

Question 1.26: ( from ncert p 27)
If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how
many volumes of water vapour would be produced?
Answer 1.26:
Reaction of dihydrogen with dioxygen can be written as:
Now, two volumes of dihydrogen react with one volume of dihydrogen to
produce two volumes of water vapour. 2H2+O2 = 2H2O (g)
Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to
produce ten volumes of water vapour.

Avogadro's Law :
Equal volumes of all
gases at the same
temperature and
pressure (sameT& P)
should contain equal
number of molecules’.

Answer 1.10:
(i)1 mole of C2H6 contains 2 moles of carbon atoms. Number of moles of
carbon atoms in 3 moles of C2H6 = 2 × 3 = 6
(ii)1 mole of C2H6 contains 6 moles of hydrogen atoms. Number of moles of carbon
atoms in 3 moles of C2H6 =3×6 = 18
(iii)1 mole of C2H6 contains 6.023 × 10 23 molecules of ethane. Number of
molecules in 3 moles of C2H6 = 3 × 6.023 × 10 23
= 18.069 × 10 23
* Numerical related to law –
ASSIGNMENT NO 6 – (Question 1.10: ncert)
In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane. ( Avogadrov’s law)

*PERIOD 7
Dalton’s atomic theory
Atomic and molecular masses
Average atomic mass
Molecular mass
Formula mass
Learning Objectives

* Calculation of value for amu
*The average atomic mass of an
element is the sum of the
masses of its isotopes, each
multiplied by its natural
abundance (the decimal
associated with percent of
atoms of that element that are
of a given isotope).

* Molecular mass and Formula mass
*Calculation of molecular formula –
molecular mass of H2O is
*= (2xatomic mass of hydrogen) + ( 1 x atomic
mass of oxygen) = 2 x 1.0008 u +16.00 u = 18 .02 u
*Calculation of formula mass –

* Assignment no 7 ( Q.1.1 ncert P-25)
* Question 1.1:
* Calculate the molecular mass of the following:
* (i) H2O (ii) CO2 (iii) CH4
* Answer 1.1:
* (i) H2O:
* The molecular mass of water, H2O
* = (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
* = [2(1.0084) + 1(16.00 u)]
* = 2.016 u + 16.00 u
* = 18.016
* = 18.02 u
* (ii) CO2:
* The molecular mass of carbon dioxide, CO2
* = (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
* = [1(12.011 u) + 2 (16.00 u)]
* = 12.011 u + 32.00 u
* = 44.01 u
* (iii) CH4:
* The molecular mass of methane, CH4
* = (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
* = [1(12.011 u) + 4 (1.008 u)]
* = 12.011 u + 4.032 u
* = 16.043 u

How to calculate of isotopic mass ?
Q. Use the data given in the following table to calculate the molar mass of
naturally occurring argon isotopes. Q. 1.32:( from ncert P - 28)
Isotope Isotopic molar mass Abundance
36 Ar 35.96755 gmol–1 0.337%
38 Ar 37.96272 gmol–1 0.063%
40 Ar 39.9624 gmol–1 99.600%
* Answer 1.32: Molar mass of argon
Ans = 39.947 gmol–1

*Period 8
*LEARNING OBJECTIVE
Mole concept and Molar mass

*Definition- One mole is the amount of a substance which
contains as many particles ( atoms/molecules/ions) as there
are atoms in exactly 12 g of 12C isotope.

*Assignment no.8
*Question 1.4: ncert P 25
*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.
*Answer 1.5:
*The balanced reaction of combustion of carbon can be written as:
*(i)As per the balanced equation, 1 mole of carbon burns in1 mole of
dioxygen (air) to produce1 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.

Question 1.7: from ncert P 26
How much copper can be obtained from 100 g of copper
sulphate (CuSO4)?
Answer 1.7:
1 mole of CuSO4 contains 1 mole of copper.
Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00
= 159.5 g
159.5 g of CuSO4 contains 63.5 g of copper.
⇒ 100 g of CuSO4 will contain of copper.
Amount of copper that can be obtained from 100 g CuSO4
= 39.81 g

Question 1.8:
Determine the molecular formula of an oxide of iron in which the mass per cent of iron and
oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69 g mol–
1.
Answer 1.8:
Mass percent of iron (Fe) = 69.9% (Given) Mass percent of oxygen (O) = 30.1% (Given)
Number of moles of iron present in the oxide = 1.25
Number of moles of oxygen present in the oxide
= 1.88
Ratio of iron to oxygen in the oxide,
= 1 : 1.5
= 2 : 3
The empirical formula of the oxide is Fe2O3.
Empirical formula mass of Fe2O3 = [2(55.85) + 3(16.00)] g Molar mass of Fe2O3 = 159.69 g
Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus,
the empirical formula of the given oxide is Fe2O3 and n is 1.
Hence, the molecular formula of the oxide is Fe2O3.

Question 1.10: ( from ncert p27)
In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Answer 1.10:
(i) 1 mole of C2H6 contains 2 moles of carbon atoms. Number of
moles of carbon atoms in 3 moles of C2H6 = 2 × 3 = 6
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms. Number of
moles of carbon atoms in 3 moles of C2H6
= 3 × 6 = 18
(iii) 1 mole of C2H6 contains 6.023 × 1023 molecules of ethane.
Number of molecules in 3 moles of C2H6
= 3 × 6.023 × 1023 = 18.069 × 1023

Learning objective
A. Percentage composition
B. Concept of empirical formula
C. Concept of molecular formula
Period 9

* What do you mean by percent composition ?

* Example - How to calculate mass percent ?

* Assignment no 9.1 ( Try your self )
*Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).
Question 1.2: ( NCERT)
Solution –Answer 1.2:
The molecular formula of sodium sulphate is Na2SO4
Molar mass of Na2SO4 = [(2 × 23.0) + (32.066) + 4 (16.00)] = 142.066 g
Mass percent of element
Mass percent of sodium: 46 X100 = 32.379 = 32.4 %
142.066
Mass percent of sulphur: = 32.066 X100 = 22.57 %
142.066
Mass percent of oxygen = 64.0 X 100 = 45.05 %
142.066

WHAT DO YOU MEAN BY EMPIRICAL AND MOLECULAR FORMULA ?

* How are empirical and molecular formulas related?

*Example - How to calculate empirical formula?

* Example - How to calculate molecular formula?
a. Q. A compound contains 86.88% of carbon and 13.12% of hydrogen, and the molecular
weight is 345. What will be the empirical and molecular formula of the compound?
b. Solution –
c. Step 1 – Multiply percent composition with the molecular weight--
d. Carbon : 345 x 0.8688 = 299.736 , Hydrogen : 345 x 0.1312 = 45.264
e. Step 2 –Divide each value by the atomic weight --
f. Carbon : 96.0852 / 12.011 = 7.9997 , Hydrogen : 10.07846 / 1.008 = 9.998
g. Oxygen : 32.0025 / 15.9994 = 2.000 , Nitrogen : 56.0238 / 14.0067 = 3.9997
h. Step 3 -Round off the values to closest whole number—
i. Carbon = 8 , Hydrogen = 10 , Oxygen = 2 , Nitrogen = 4
j. Hence, the molecular formula is C8H10N4O2.
k. Step 4 - Since 2 is the common factor among 8, 10, 4 and 2.
l. The empirical formula is C4H5N2O.

* Assignment 9.2 ( try yourself)
* Q. A compound contains 86.88% of carbon and 13.12% of hydrogen, and
the molecular weight is 345. What will be the empirical and molecular
formula of the compound?
* Solution –
*Step 1 - Carbon : 345 x 0.8688 = 299.736 , Hydrogen : 345 x 0.1312 = 45.264
*Step 2 - Carbon : 299.736 / 12.011 = 24.995 , Hydrogen : 45.264 / 1.008 = 44.91
* Step 3 - Molecular formula is C25H45
*Step 4 -Since 5 is the common factor, the empirical formula is C5H9

*Perriod10
*Learning objective
 Stoichiometry and calculations
Concept of limiting reagent
Reactions in solutions and methods to express concentration of solutions.

*What do you mean by Stoichiometry
*Definition - Stoichiometry is an important concept in
chemistry that helps us use balanced chemical equations to
calculate amounts of reactants and products.
*Stoichiometry helps us determine how much substance is
needed or is present. Things that can be measured are;
Reactants’ and Products mass
Molecular weight
Chemical equations
Formulas

* What do you mean limiting reagent and excess reagent ?
* Limiting reagent “ The
reactant which reacts
completely in the reaction is
called limiting reactant or
limiting reagent. ”
* Excess reagent – “ The
reactant which is not
consumed completely in the
reaction is called excess
reactant . ”

* Examle- How to carry on calculations in stoichiometry ?
*Question 1.24: (ncert)
*Dinitrogen and dihydrogen react with each other to produce ammonia
according to the following chemical equation:
*N2(g) + H2(g) → 2NH3(g)
*(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen
reacts with 1.00
*× 103 g of dihydrogen.
*(ii) Will any of the two reactants remain unreacted?
*(iii) If yes, which one and what would be its mass?
*Solution in next slide

*Solution Q.1.24 (ncert)
*(i)Balancing the given chemical equation,
*N2(g) + 3H2(g) → 2NH3(g)
*From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2
mole
(34 g) of ammonia.
*⇒ 2.00 × 103 g of dinitrogen will react with 6 g X 2.00 X 10 3 dihydrogen i.e., 2.00 × 103 g
* 28
of dinitrogen will react with 428.6 g of dihydrogen.
*Given,
*Amount of dihydrogen = 1.00 × 103 g Hence, N2 is the limiting reagent.
*28 g of N2 produces 34 g of NH3.
*Hence, mass of ammonia produced by 2000 g of N2 = 38 x 2000g = 2428.57 g
* 28 g
*(ii) Hence - N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will
remain unreacted.

Assignment 10 ( try at home)
*Question 1.23: ( ncert)
*In a reaction A + B2 → AB2
*Identify the limiting reagent, if any, in the following reaction mixtures.
*(i) 300 atoms of A + 200 molecules of B
*(ii) 2 mol A + 3 mol B
*(iii) 100 atoms of A + 100 molecules of B
*(iv) 5 mol A + 2.5 mol B
*(v) 2.5 mol A + 5 mol B

*Solution of assignment – 10.1
*Answer 1.23:
*A limiting reagent determines the extent of a reaction. It is the reactant which is the
first to get consumed during a reaction, thereby causing the reaction to stop and limiting
the amount of products formed.
*(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200
molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused.
Hence, B is the limiting reagent.
*(ii) According to the reaction, 1 mole of A reacts with 1 mole of B. Thus, 2 mole of A will
react with only 2 mole of B. As a result, 1 mole of A will not be consumed. Hence, A is the
limiting reagent.
*(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all
100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is
stoichiometric where no limiting reagent is present.
*(iv) 1 mole of atom A combines with 1 mole of molecule B. Thus, 2.5 mole of B will
combine with only 2.5 mole of A. As a result, 2.5 mole of A will be left as such. Hence, B is
the limiting reagent.

* Method of expressing concentration of solutions in reactions
*Methods of Expressing Concentration of Solutions
*Mass percent or weight per cent ( w/w % )
*Mole fraction (x)
*Molarity (M)
*Molality (m)
*Normality (N)

* 1. The mass percent -The Mass per cent formula is expressed as
solving for the molar mass also for the mass of each element in 1mole
of the compound.
Mass percent = mass of the solut x 100 %
mass of the solution
*2. The Mole fraction – (x) Mole fraction is a unit of
concentration, defined to be equal to the number of moles
of a component divided by the total number of moles of a
solution.
mole fraction of solute A is = (moles of A) ÷ (total moles),
mole fraction of the solvent = (moles of solvent) ÷ (total moles).

*Example – calculation of mass percent
* Q. An aqueous solution of sodium chloride (CaCl2) is prepared.
Determine the mass of 5%(m/m) solution of calcium chloride that can be
prepared using 100 g of CaCl2’
*Solution –
*The interpretation of m/m % number shows that 5g of CaCl2 is used to
prepare 100g of solution.
*Therefore,
*g solution = (100g CaCl2 x 100g solution) / 5g CaCl2
*g solution = 2 x 103 g solution

*What do you mean by molarity, molality and normality?
*Definition-
*Molarity- (M) – The most wisely used unit od concentration is defined as
– “ The number of moles of the solute present in 1 liter of solution .”
*Molality – (m)” The number of moles of the solute present in 1 kg of
solvent in a solution.”
*Normality-(N) – “ Defined as the number of mole equivalents (gram
equivalents) per liter of solution. ”

*Differene between molarity and molality

*Some solved examples related
*Question 1.25: ( ncert)
*How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
*Answer 1.25:
*Molar mass of Na2CO3 = (2 × 23) + 12.00 + (3× 16) = 106 g
mol–1 Now, 1 mole of Na2CO3 means 106 g of Na2CO3.
*0.5 mol of Na2CO3
*= 53 g Na2CO3
*⇒ 0.50 M of Na2CO3 = 0.50 mol/L Na2CO3
*Hence, 0.50 mol of Na2CO3 is present in 1 L of water or 53
g of Na2CO3 is present in 1 L of water.

*Solved example of normality
*Q. Calculate the normality of 0.321 g sodium carbonate when it is
mixed in a 250 mL solution.
*Solution:
*First, you have to know or write down the formula for sodium
carbonate. Once you do this you can identify that there are two
sodium ions for each carbonate ion. Now solving the problem will
be easy.
*N of 0.321 g sodium carbonate
*N = Na2CO3 × (1 mol/105.99 g) × (2 eq/1 mol)
*N = 0.1886 eq/0.2500 L
*N = 0.0755 N

*SOLVED EXAPLE OF MASS FRACTION
*
*Solution
*Moles of CH3OH = 5.5 / 32 = 0.17 mole
*Moles of H2O = 40 / 18 = 2.2 moles
*Therefore, according to the equation
*mole fraction of CH3OH = 0.17 / 2.2 + 0.17
*mole fraction of CH3OH = 0.073
Q. Determine the mole fraction of CH3OH and H2O in a
solution prepared by dissolving 5.5 g of alcohol in 40 g of
H2O. M of H2O is 18 and M of CH3OH is 32.

*Question 1.28: (ncert)
*Which one of the following will have largest number of atoms?
*(i) 1 g Au (s) , (ii) 1 g Na (s) , (iii) 1 g Li (s) , (iv) 1 g of Cl2(g)
*Answers- Lets calculate no of atoms in each case -
*(i) 1 g of Au (s) = 1/197 mol of Au (s) = 6.023x 1023
/197 atoms of Au (s)
* = 3.06 × 1021
atoms of Au (s)
*(ii) 1 g of Na (s) = 1/23 mol of Na (s) = 6.023x 1023/197 atoms of Na (s)
*= 0.262 × 1023 atoms of Na (s) = 26.2 × 1021
atoms of Na (s)
*(iii) 1 g of Li (s) = 1/7 mol of Li (s) = 6.023x 1023/197 atoms of Li (s)
*= 0.86 × 1023 atoms of Li (s) = 86.0 × 1021
atoms of Li (s)
*(iv) 1 g of Cl2 (g) = 1/ 71 mol of Cl2 (g) as
*(Molar mass of Cl2 molecule = 35.5 × 2 = 71 g mol–1) atoms of Cl2 (g)
*= 0.0848 × 1023 atoms of Cl2 (g) = 8.48 × 1021 atoms of Cl2 (g)
*Hence, 1 g of Li (s) will have the largest number of atoms.
*Assignment no 10.2

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