Docsity
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Docsity AI
Docsity AI
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Mole concept Handwriting simplified notes, Summaries of Chemistry

Chemistry

These handwritten NEET notes are based on NCERT and are useful for quick revision. Includes important concepts, diagrams, key points, and exam-oriented explanations. Helpful for Class 11/12 and NEET preparation

Typology: Summaries

2019/2020

Available from 05/07/2026

abhi-gyaan
abhi-gyaan 🇮🇳

1 document

1 / 22

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
some
basic
Concept
of
Chemistry
:
-
Physical
Property
:
-
The
Property
which
can
be
measured
without
changing
the
Chemical
Composition
of
the
substance
is
known
as
Physical
Property
like
mass
,
volume
,
density
etc
.
Chemical
Property
:
-
The
property
which
Can
be
evaluated
at
the
cost
of
matter
itself
is
known
as
chemical
property
.
Classification
of
Matter
:
-
Matter
V
V
Pure
Substance
Mixture
#
I
1
↓
↓
Element
Compound
2/
Y
Homogenous
Heteroge-
nous
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16

Partial preview of the text

Download Mole concept Handwriting simplified notes and more Summaries Chemistry in PDF only on Docsity!

some basic^ Concept

of Chemistry :

Physical Property^ : -

The

Property

which can^ be measured without

changing the Chemical^ Composition (^) of the

substance is known^ as

Physical (^) Property

like mass^ , volume^ ,

density etc. Chemical Property : -

The

property

which Can be^ evaluated at the

cost (^) of matter (^) itself is known (^) as chemical property . Classification of^ Matter^ :^

Matter

V (^) V

Pure Substance Mixture

I 1
Element Compound

2/ Y

Homogenous Heteroge-

nous

Law ofChemical^ Combination :^ -

Law of Conservation of Mass^

: - For

any chemical change

total mass

of active^ reactants^ are always equal^

to the mass of the Product

Formed.^ It^ is^ a^ derivation^ of Dalton's atomic theory. ' Atoms neither (^) created nor destroyed

  • &

Total masses of reactants= Total^ masses of

Products +^ masses^ of unreacted reactants - Example:^ -4. 90g of KCO3^ When^ heated^ Produced 1 :

92g of^ oxygen

and the^ residure^ (ke) Left

behind weights

96g

. Show (^) that (^) these results

illustrate the^ law^

of Conservation^ of mass . Soli-weight of KCI0z taken^ = (^) u. gog total weight of^ the (^) Products (^) (KC +

= 2.^96 +^1. 92 =^4. 88g Difference between the wight of^ the reactant^ and the

total

weight of^ the Product =^4.^ 90-4.^ 88-^. 02g

This small^ difference

may

be (^) due to (^) experimental error. Thus (^) , Law

of

Conservation of mass^ holds

good

within (^) experimental error.

Example :^ -Ratio^ of masses (^) of Carbon^ and (^) Sulphur which Combine^ with^ Fixed^ mass^ (32 (^) Part) (^) of

Oxygen

is 12 :^32 or 3 :^ 8. ..... ()

In CS2 ratio of masses

of Carbon (^) and (^) Sulphur is 12 :^64 or 3 :^16 ..... (2) So (^) & M The two^ ratio (^) (1) and (^) (2) are related to each other by (^) - :^ % or^2 :^1 or 3 : 2(5) First ratio (^) is integral multiple^ of^

second.

Gay

Lussac's Law^ of

Combining

Volumes :^ -

At given

temperature and^ pressure the

volume of

all

gaseous

reaction and products

bear (^) a (^) simple whole (^) number (^) ratio to each

other.

Example:^

  • H2(g) + <12(g) - > 2HC(g) ( : (^2) : 2) One (^) Volume (^) of hydrogen reacts with (^) one (^) volume of chlorine^ to^ form^ two^ volume^ of Hel^ gas

the (^) ratio by

Volume which^ gases bear^ is^1 :^ L^ :^2

which is a (^) simple whole^ number (^) ratio. Other Example are^ :

25a +^02

= 2503[ : 1 : (^) 2] (^2) Volume (^1) Volume 2 volume 2NH

= N2 + 3H2[2 :^1 :^ 3]

(^2) volume ↓ (^) Volume (^3)

Volume

NH3 +^ HCl^

NHy (l[Vapour]^ [^ : (^2) : (^) 1] I (^) volume (^) (Volume L volume Dalton's Atomic Theory :

In (^1808) , Dalton^ Published^ 'A^ new^ system of

chemical

philosophy:

in which^ he^ proposed

the following

:

=> Matter Consist^ of indivisible atoms.

  • > All the atoms (^) of a

given

element have^ identical

Properties including

identical mass. Atoms of

different elements^ differ in^ mass.

  • >

Compound are^ formed^ when^ atom^ of different

element Combined in a fixed Ratio.

One atom^ of^ the element^ is heavier^ in^ Comparision

to the mass of one atom of

hydrogen.

Atomic mass of an element : -

=> Mass of^ one^ atom^ of^ the^ element

Mass of^ th Part of mass of C-12^ atom

  • (^) Average Atomic^ mass^ : - Elements (^) are found in (^) different isotopic Forms (atoms (^) of same^ element^ having different atomic mass)

So the atomic mass of

any element is the average of^

all the isotopic mass within

a given

sample.

Average atomic^ mass^

= 22 % abundance) X atomic mass

Mole (^) Concept :^ -Mole is^ the number (^) equal to

Avogadro's

number just like a

doyen

is

equal

to (^12) , a Century means too, (Na = 6. 022 x (^) 1023)

6 : 02x^

  1. (^4) (of gas at^ NTP O
Particles or^ Molar^ Vol^.

In terms^ In^ terms^ of of particles ↓^ mole^ Volume In (^) terms (^) of mass

L ↓ gram gram^

formula

Igram e of

mole of mass of Substance

No. of mole (^) of Molecule =Witinemass No. of Moles (^) of atoms =lgtina No. (^) of Moles^ of Gases =oatSe volume (Standard Molar^ Volume^ at^ STP^ =^22.^4 lit)

Example: -^ Molecular^ mass^ of Calcium^ Nitrate^ :

Ca(NO3)2 =^ atomic^ mass^ of Ca^ +^ 2X^ Atomic

Mass (^) of N +^6 X^ Atomic mass of 0 =^40 +^ 2x14 + 6416 = 40 + (^28) + 96 = (^) 164u % of Calcium^ ((a) =MassOfC100 = xoO 24 · 39

% of Nitrogen (N) =^ Mass^ of N

Molecular mass^ of Ca(NO) 100 = x00 =^17 % of

Oxygen (0)^

=SS^ Ot^ of Ca(NO)X100 = x Example :^ How much^ of (^) Copper is^ obtained from

    • Loog of^

Copper Sulphate^ (Cuson)?

Sol :^ -Molecular mass

of Cusoy^

= Atomic Mass of Cu +

atomic mass (^) of Stux (^) atomic mass (^) of 0 = 63.^5 + (^32) + (^) 4X = (^159). 5 y Gram (^) Molecular mass

of Cuson^

= 159. 59

Now , 159.^5

g of^

Cuson here^ C^ =^63.

5g

: 100g of^

CuSOy have^ C^ =^63.

5 X = 39. g

Empirical Formula (^) :

It (^) is the Formula^ which (^) expresses the (^) smallest whole number ratio of

the Constituent atom

within the Molecule.

Ex :^ -^ A^ Compound Contains C^ =^71.^23 %, H^ =^12.^95 %

and 0 =^15.^81 %.^ What is^ the^ empirical Formula

of the^ Compound. Sol!^ - Element's Atomic^ Weight in^ Relative Simplest Empirical Name with mass (^) Compound No·^ of atomic (^) Formula their (^) Symbol Atom^ Ratio

Carbon (c)^12

. 93

93693- 12.^95

  1. 988

I

Hydrogen

(H) 1 12.^95

T = 12.^95 295 :^13 Cultis O -Mygen(o)

(^818090098) Chemical Equation and Stoichiometry

Symbol:

  • Liquid - > (1) Gas -^ >^ (g)

(3) - Balance the (^0) atom (^) :^ -

CazNa +^ GH^
  • >

3 (a(OH)2 +^ 2NHs

upon checking

, the^ H^ atoms^ are^ already balanced

: (^) Final (^) balanced chemical equation is^ :

CasN2+^ 6H

-- >

3 CalOH)2 +^ 2 NH

Balance the following (^) by (^) yourself. & (a^ +^ Ho -^ (a(OH)2 +^ Hz ② Aly^ +^ H^

  • AloH + Hy

* Problem^

Involving

mass-mass Relationship :

Example :^ What^ mass^ of Copper oxide^ will^ be^ obtained by Catomic maneating1235g^ of Copper Carbonata

Sol :^ - The chemical equ for the reaction is!^ -

Cu(O

eat (^) >

Cu0 +^ CO
63.^5 +^12 +^48 63.^5 +^16

59 => 79. 59

5g

of Copper Carbonate^ upon

heating (^) gives C = 5 d.og of^ Copper^

Carbonate

upon (^) heating given (^) Cu : (^12). 35g of (^) Copper Carbonate^ upon heating (^) give Cy = = x12. 35 =

9

Problem Involving Mass-Volume (^) Relationship!-

  • Example :^ - KClOz On heating decompose^

to

give

KCI and Of What is^ the volume

of O at NTP (^) Liberated by 0.^2 mole of Keloz^ ? sol": · (^) The (^) chemical

equ

for the decomposition of KCOz

is

Heat

2k(

  • > 2kC +

(^2) mol 3 Mol (3x22.^ 4h^ =^67 · 2L) 2 mole of KC^

evolve O2 at NTP = 67. 21

↓ (^) mole of K2103 evolve^ O2 at^ NTP^ = Ge 0 .I mole^ of KC03 evolve (^) O2 at (^) NTP (^) =1.^2 2X

. /1= 3 .36L Problem Involving Volume-volume Relation Ship : - Example:

↓ Litre of

Oxygen at NTP is^ allowed to react with (^3) Liter (^) of Carbon monoxide at NTP·^ Calculate^ the^ volume of each

gas

found

after the^ reaction^.

Eg

. (^) 30g of He^ react^ with^29 . 09 O to^ yield t ① What (^) is limiting reactant .?. ② calculate^ the^ maximum^ amount^ of H20 that Can be (^) formed ③ Calculate^ the^ amount^ of one^ of the reactant which remains unreacted. sol. The balanced Chemical

equ

is

2H2 +^02

  • 2420 2x2 = ug 32g^ 2x18 =

36g

Step-

  • Determination of limiting reactant

4g of^ H^

react with^02 =

32g

: 3g of^ He^ react with^02 = 2x = 24g

but the amount

of oxygen

actually available = 29g As

Oxygen

is available in excess ,

therefore , hydrogen is the limiting reactant.

Step-1-Calculation of the^ amount^ of water

thatCan^ be^ formed. 4g of (^) H2 From^ H20 in^ the^ reaction^ = 369 :

3g

of He from^ to^ in^ the^ reaction^ =^ 36 ↑ X 3 = 279

Step

-> Calculation of the amount

of

oxygen

left unreacted.

4g ofHe required o^ for (^) reaction = 329 :. 3g of He required^ For reaction^ = x3 = zug but Oxygen

which

actually Present-29g : (^) Amount of (^) Oxygen Left^ unreacted^ = 29-245g

  • (^) Method of Expressing Concentration (^) of

Solution : -

Molarity (M)

: - The

Molarity of a

Solh is the no of moles

of solutePresent^ in^ one^ litre^ (1dm3)^ of the^ Som^ . M = , (^000) me M =^ No .

of moles^ ofsolute

volume of Soln^ (in^ litre) w =^ mass^ of solute (^) , M (^) , = (^) Molar (^) mass of solute v =^ volume of Solution.

It density of^ the solution^ is^ dyles , mass of

solution=

grams .

mass (^) of solute^ = MMLg[M is molar mass^ of solutel

mass of Solvent^ =

1000d-MM2g Molality (m) (^) = MM

X 1000

=o

MM

m

2/ ( or m (d- = MormM2) or M = (^2) /1 000

  • (^) Relationship between^ Molality (m) and mole (^) Fraction (^) (Xc) :^ - Molality(m) = moles^ ofSolute/mass ofSolvent Linkg)

Mole Fraction of Solute^ (X 1 ) =^ Moles^ of solute^ /(mole of solute^ +^ moles^ of

a^ Solvent)

  • n. =^ Mole

of

solute We =^ mass of (^) Solvent(g) n2 =^ mole^ of solvent S m = molality of solute M2 =^ Molar mass of Solvent(ymos)^ X1 =^ Mole (^) Fraction of solute From molality : m =

W 12 = Now ,^ mole^ Fraction^ of^ solute :

=

Final

Relationship :-

: or inversely m Relationship between^ Molavity (m) (^) and mole Fraction (*)

Mole Fraction . X

= -MMMM