Download 5. gram molecular gram molecular mass 1. Modern atomic theory 3. i

CLASS X
(answer key) chemistry
2. Atoms and Molecules
Part - A
6.
1. Isotopes - 17Cl35 , 17Cl37 Isobars - 18Ar40,20Ar40
Molecular mass
28
2. Atomicity of N2 = ----------------------- = --- = 2
Atomic mass
14
given particles
3.01151 * 1023
No. of moles = --------------------- = ------------------ = 0.5
mole.
6.023*1023
6.023 2*1023
3. Gram molar
Gram molar mass of O2
32g
volume of O2 = ----------------------------- = -----Density of O2
1.429g/cc
7. (ii) Isotones
8.
Given
Hint Density= mass/volume
V=M/D
i. Chlorine
ii. Neon
iii.Phosphorous
iv.Ozone
4.
Atom
The smallest particle of an
element that can take part
in a chemical reaction
an atom is a non bonded
entity.
An atom may or may not
exist freely.
June/15
= 22.4 L
Molecules
The smallest particle of an
element or a compound that
can exist freely.
A molecule is a bonded
entity.
An molecule can exist freely
Cl2
Ne
P4
O3
No. of atoms
per molecule
2
1
4
3
Atomicity
Diatomic
Monotomic
Polyatomic
Triatomic
9. i. The Molar Volume of gas at STP is 22.4 liters
ii. 2*vapour density = Relative molecular Mass.
iii. An atom may or maynot exist independently.
iv. The ratio of atoms in a molecule may be fixed and
integral but may not be simple.
V.H2O is a hetero atomic molecule.
5. gram molecular mass of H20
2(H) = 2*1 = 02
1(O) = 1*16 = 16
----gram molecular mass of H20
18 g
-------
10.
i. One Mole of Avogadro number
ii. Gram molar volume
iii. One atomic mass unit (1 amu)
iv. Vapour density V. Atomicity
.
PART- B
1. Modern atomic theory
An atom is the smallest particle which takes part in chemical
reaction.
An atom is considered to be a divisible particle.
The atoms of the same element may not be similar in all
respects. eg: Isotopes - 17Cl35 , 17Cl37
The atoms of different elements may be similar in some
respects. eg: Isobars - 18Ar40,20Ar40
The ratio of atoms in a molecule may be fixed and integral
but may not be simple. e.g., C12H22O11 is not a simple ratio.
(Sucrose)
Atoms of one element can be changed into atoms of other
element by transmutation.
The mass of an atom can be converted into energy. This is in
accordance with Einstein’s equation E =mc2
2.
Relative. Molecular Mass (RMM):
It is defined as the ratio of the mass of 1 molecule of the gas
or vapour to the mass of 1 atom of hydrogen
Mass of 1 molecule of the gas or vapour
Relative molecular = ---------------------------------------------------mass of a gas
Mass of 1 atom of hydrogen
Vapour. Density (V.D): It is defined as the ratio of the mass
of a certain volume of the gas or vapour to the mass of the
same volume of hydrogen at the same temperature and
pressure.
3. i.
Given particles 212.046 *1023
No. of
= ----------------= -------------= 2 moles
23
moles in Cu
6.023*1023
1 6.023 *10
iii No. of moles
Given particles 1.51*1023
Mass of 1 volume of gas or vapour
V.D = -------------------------------------------------Mass of 1 volume of hydrogen
Applying Avogadro’s law
Mass of 1 molecule of gas or vapour
V.D = --------------------------------------------------Mass of 1 molecule of hydrogen
Since hydrogen is diatomic,
V.D =
Mass of 1 molecule of gas or vapour
-----------------------------------------------------2 x Mass of 1 atom of hydrogen
Multiplying both side by 2 , we get
Mass of 1 molecule of gas or vapour
2* V.D = ----------------------------------------------------* 2
2 * Mass of 1 atom of hydrogen
Mass of 1 molecule of gas or vapour
2* V.D = ---------------------------------------------------Mass of 1 atom of hydrogen
2* V.D = Relative molecular mass of a gas or vapour
2* Vapour Density = Relative molecular mass
ii.
Given mass
No .of
= - ------------moles in Fe
Atomic mass
27.95 1
= -- ---------55.85 2
= 0.5 moles
5
in CO2
= ------------------- = ------------6.023*1023
6.023*1023
= 0.25 moles
4. . i. gram molecular mass of H20
2(H) = 2*1 = 02
1(O) = 1*16 = 16
gram molecular mass of H20 18 g
ii. gram molecular mass of CO2
1(C) = 1*12 = 12
2(O) = 2*16 = 32
gram molecular mass of CO2 44g
iii. gram molecular mass of NaOH
1(Na) = 1*23 = 23
1(O) = 1*16 = 16
1(H) = 1*1 = 01
gram molecular mass of NaOH
40g
iv. gram molecular mass of NO2
1(N) = 1*14 = 14
2(O) = 2*16 = 32
gram molecular mass of NO2 46g
v. gram molecular mass of H2SO4
2(H) = 2*1 = 02
1(S) = 1*32 = 32
4(O) = 4*16 = 64
gram molecular mass of H2SO4 98g
Element
Atomic mass Molecular mass
Atomicity
Chlorine
35.5
71
(71/35.5)=2
Ozone
(48/3)=16
48
3
Sulphur
32
(32*8)=256
8
6. No. of molecules
Given mass
0.18 0.01
= ------------------ * 6.023*1023 = --------* 6.023*1023
Molecular mass
18
= 0.01 * 6.023*1023 = 6.023*1021 H2O molecules
7.i. 1 mole of Ca (40g) and 1 mole of oxygen atom (16g)
combine to form 1 mole of CaO (56g)
ii. 1 mole of Ca (40g) and 1 mole of C (12g) and 3 moles of
oxygen atom (48g) ) combine to form 1 mole of CaCO3 (100g)
8. I. 5 mole of water ( H20 )
2(H) = 2*1 = 02
Mass = molecular mass * no. of moles 1(O) = 1*16 = 16
= 18 * 5 = 90g
GMM of H20 18g
ii. 2 moles of Ammonia (NH3)
1(N) = 1*14 = 14
Mass = molecular mass * no. of moles
3(H) = 3*01 = 03
= 17 * 2 = 34g
GMM of NH3 17g
iii. 2 moles of Glucose (C6H12O6)
06(C) = 06*12 = 72
Mass = molecular mass * no. of moles 12(H) = 12*01 = 12
= 180 * 2 = 360g
06(O) = 06*16 = 96
GMM of (C6H12O6) = 180g
Part -C
1 i. Given Volume of NH3 = 67.2L ; No. moles of NH3 = ? 5,i. mass of chlorine (Cl2 )= molecular mass*no.of moles
We know,
22.4 L of any gas = 1 mole
= (2*35.5)*1
volume of NH3
67.2 L
= 71g
No. of moles = -------------------- =
---------= 3 moles
of NH3
22.4 L
22.4 L
ii. mass of sulphur (S8)= molecular mass*no.of moles
ii. Mass of NH4Cl
ii. Ammonia gas (NH3)
= (8*32)*2 = 256*2
01(N) = 01*14
= 14
= 512g
04(H) = 04*01 = 04
iii. NH3 +HCL------NH4Cl
iii. mass of Ozone (O3) = molecular mass*no.of moles
01(Cl) = 01*35.5 = 35.5
= ( 3*16)*4 = 48*4
Mass of NH4Cl
= 53.5g
= 192g
2.i. a, Nitroglycerine = 1mole
iv. mass of Nitrogen (N3)= molecular mass*no.of moles
b, No. of moles of gas molecules = 19
= ( 2*14)*2 = 28*2
ii.CO2 = 12 moles iii. Mass of 1 mole of nitroglycerine
= 56g
N2 = 6 moles
03(C) = 03*12
= 36
O2 = 1 moles
05(H) = 05*01
= 05
6. i.
Given mass
2
Gas
= 19
03(N) = 03*14
= 42
No. of moles of nitrogen =--------------= ------ = 0.14mole
molecules
09(O) = 09*16
= 144
atomic mass 14
Mass of C3H5((NO3))3 =227g
3. i. 2 moles of sodium bi carbonate ii. Mass of NaHCO3
ii.
Given mass
23
iii.1 moles of CO2
1(Na) = 1*23 = 23
No. of moles of sodium = ------------- -= ------ = 1 mole
1(H) = 1*01 = 01
atomic mass
23
1(C) = 1*12 = 12
3(O) = 3*16 = 48
iii.
Given mass
40
Mass = 104g
No. of moles of calcium = ------------- -= ------ = 1 mole
4.i .mass of CaO =174g ; mass of Ca = 100g
atomic mass
40
Mass of Oxygen = 174-100 =74g
ii.
mass of oxygen 74
No.of moles in oxygen = -------------------- = ---- = 4. 6moles
iv.
Given mass
1.4
atomic mass
16
No. of moles of lithium = ------------- -= ------ = 0.2 mole
atomic mass
7
iii.
mass of calcium 100
No.of moles in calcium = -------------------- = ------ = 2.5 moles
v.
Given mass
32
atomic mass
40
No. of moles of sulphur = ------------- -= ------ = 1 mole
iv. mass of calcium in 174g of calcium oxide = 100g
atomic mass 32
mass of calcium in 1000g of CaO= (100/174)*1000 = 574.71g
Prepared by
Mr. Johnson Prabhu, M.Sc.,B.Ed.
St.Paul’s Mat. Hr. Sec. School,
Block-4, Neyveli - 607801