Class 11 Chemistry Chapter 1 Question Answers . Some Basic Concepts of Chemistry Class 11 Notes .
Some Basic Concepts of Chemistry
1. 1
Calculate the molecular mass of
the following :
(i) H2O
(ii) CO2
(iii) CH4
Solution
:
(i) H2O
Molecular weight of water, H2O
= (2 x Atomic
weight of hydrogen) + (1 x Atomic weight of oxygen)
= [2(1.0084) + 1(16.00
u)]
= 2.016 u +16.00 u
= 18.016u
So approximately
= 18.02 u
(iii) CO2
= Molecular weight
of carbon dioxide, CO2
= (1 x Atomic
weight of carbon) + (2 x Atomic weight of oxygen)
= [1(12.011 u) +
2(16.00 u)]
= 12.011 u +32.00
u
= 44.011 u
So approximately
= 44.01u
(ii) CH4
Molecular weight
of methane, CH4
= (1 x Atomic
weight of carbon) + (4 x Atomic weight of hydrogen)
= [1(12.011 u) +4
(1.008u)]
= 12.011u + 4.032
u
= 16.043 u
1.2 Calculate the mass percent of
different elements present in Sodium Sulphate (Na2SO4).
Solution
:
Molar mass of NaSO4
= [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 g
Mass percent of an
element = (Mass of that element in compound/Molar mass of that compound) × 100
∴ Mass percent of
sodium (Na): (46/142) × 100 = 32.39%
Mass percent of
sulphur(S): (32/142) × 100 = 22.54%
Mass percent of
oxygen:(O): (64/142) × 100 = 45.07%
1.3 Determine the
empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen
by mass.
Solution
:
% of iron by mass
= 69.9 % [Given]
% of oxygen by
mass = 30.1 % [Given]
Atomic mass of
iron = 55.85 amu
Atomic mass of
oxygen = 16.00 amu
Relative moles of
iron in iron oxide = % mass of iron by mass/Atomic mass of iron = 69.9/55.85 =
1.25
Relative moles of
oxygen in iron oxide = % mass of oxygen by mass/Atomic mass of oxygen =
30.01/16=1.88
Simplest molar
ratio = 1.25/1.25 : 1.88/1.25 ⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula
of the iron oxide is Fe2O3 .
1.4 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.
Solution
:
The balanced
reaction of combustion of carbon in dioxygen is:
C(s) +
O2 (g) → CO2 (g)
1mole 1mole(32g) 1mole(44g)
(i) In dioxygen,
combustion is complete. Therefore 1 mole of carbon dioxide produced by burning 1 mole of carbon.
(ii) Here, oxygen
acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it
will react with 0.5 mole of carbon to give 22 g of carbon dioxide.
(iii) Here again
oxygen acts as a limiting reagent as 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.
1.5 Calculate the mass of sodium acetate (CH3COONa)
required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium
acetate is 82.0245g mol-1 .
Solution
:
0.375 M aqueous
solution of sodium acetate means that 1000 mL of solution containing 0.375
moles of sodium acetate.
∴No. of moles of
sodium acetate in 500 mL = (0.375/1000)×500 = 0.375/2 = 0.1875
Molar mass of
sodium acetate = 82.0245g mol-1
∴ Mass of sodium
acetate acquired = 0.1875×82.0245 g =
15.380g
1.6 Calculate the concentration of nitric acid in
moles per litre in a sample which has a density, 1.41 g mL-1 and the
mass per cent of nitric acid in it being 69% .
Solution
:
Mass percent of
69% means that 100g of nitric acid solution contain 69 g of nitric acid by
mass.
Molar mass of
nitric acid (HNO3) = 1 + 14 + 48 = 63g mol-1
Number of moles in
69 g of HNO3 = 69/63 moles = 1.095 moles
Volume of
100g nitric acid solution = 100/1.41 mL
= 70.92 mL = 0.07092 L
∴ Conc. of HNO3
in moles per litre = 1.095/0.07092 = 15.44 M
1.7 How much copper can be obtained from 100 g of
copper sulphate (CuSO4 )?
Solution
:
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.
∴ copper can be
obtained from 100 g of copper sulphate = (63.5/159.5)×100 = 39.81g
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
Solution
:
% of iron by mass
= 69.9 % [Given]
% of oxygen by
mass = 30.1 % [Given]
Atomic mass of
iron = 55.85 amu
Atomic mass of
oxygen = 16.00 amu
Relative moles of
iron in iron oxide = %mass of iron by mass/Atomic mass of iron =
69.9/55.85 = 1.25
Relative moles of
oxygen in iron oxide = %mass of oxygen by mass /Atomic mass of oxygen =
30.01/16 =1.88
Simplest molar
ratio = 1.25/1.25 : 1.88/1.25 ⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula
of the iron oxide is Fe2O3.
Mass of Fe2O3
= (2×55.85) + (3×16.00) = 159.7 g mol-1
n = Molar mass
/Empirical formula mass = 159.7/ 159.6 = 1(approx)
Thus, Molecular
formula is same as Empirical Formula i.e.
Fe2O3.
1.9 Calculate the atomic mass (average) of
chlorine using the following data :
% Natural Abundance Molar Mass
35Cl 75.77 34.9689
37Cl 24.23 36.9659
Solution
:
Fractional
Abundance of 35Cl = 0.7577 and Molar
mass = 34.9689
Fractional
Abundance of 37Cl = 0.2423 and Molar
mass = 36.9659
∴ Average Atomic
mass = (0.7577 × 34.9689) amu + (0.2423 × 36.9659)
= 26.4959 + 8.9568
= 35.4527
1.10 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.
Solution :
(i) 1 mole of C2H6 contains 2 moles of Carbon atoms
∴ 3 moles of of C2H6 will contain 6 moles of Carbon atoms
(ii) 1 mole of C2H6
contains 6 moles of Hydrogen atoms
∴ 3 moles of of C2H6
will contain 18 moles of Hydrogen atoms
(iii) 1 mole of
C2H6 contains Avogadro's no. 6.02 ×1023
molecules
∴ 3 moles of of C2H6 will contain ethane molecule = 3×6.02 × 1023=
18.06 ×1023 molecules.
1.11 What is
the concentration of sugar (C12H22O11) in mol
L-1 if its 20 g are dissolved
in enough water to make a final volume
up to 2L?
Solution
:
Molar mass of
sugar (C12H22O11) = (12 ×12) +(1 ×22)+ (11×16) = 342 g mol-1
No. of moles in
20g of sugar = 20/342 = 0.0585 mole
Volume of Solution
= 2L (given)
Molar
concentration = Moles of solute/Volume of solution in L = 0.0585mol /2L =
0.0293 mol L-1 = 0.0293 M
1.12 If the density of methanol is 0.793 kg L-1 ,
what is its volume needed for making 2.5 L of its 0.25 M solution ?
Solution
:
Molar mass of methanol
(CH3OH) = (1×12) + (4×1) + (1×16) = 32 g mol-1 = 0.032 kg
mol-1
Molarity of the
solution = 0.793 /0.032 = 24.78 mol L-1
Applying, M1V1
(Given Solution) = M2V2 (Solution to be prepared)
24.78×V1
= 0.25×2.5 L
V1=
0.02522 L = 25.22 mL
1.13 Pressure is determined as force per unit area
of the surface. The SI unit of pressure, pascal is as shown below :
1Pa = 1N m-2 . If mass of air at sea level is 1034 g cm-2,
calculate the pressure in Pascal .
Solution
:
Pressure is the
force (i.e. weigh) acting per unit area.
P= F/A = 1034g ×
9.8ms-2/cm2
= 1034g
× 9.8ms-2/cm2 × 1kg /1000g × 100cm /1m × 100cm /1m = 1.01332 ×105 N
Now,
1Pa = 1N m-2
∴ 1.01332 ×
105 N ×m-2= 1.01332 ×105 Pa
1.14 What is
the SI unit of mass ? How is it defined?
Solution
:
The SI unit of
mass is kilogram (kg).
The kg is defined
as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an air-tight
jar at International Bureau of Weigh and Measures in France .
1.15 Match the following prefixes with their
multiples:
Prefixes Multiples
(i)
Micro 106
(ii)
Deca 109
(iii)
Mega 10-6
(iv)
Giga 10-15
(v)
Femto 10
Ans:- micro- 10-6 , deca- 10,
mega – 106, giga – 109
, femto – 10-15
1.16. What do you mean
by significant figures ?
Ans:-
Significant
figures are meaningful digits which are known with certainty including the last
digit whose value is uncertain.
17. A sample of
drinking water was found to be severely contaminated with chloroform, CHCl3
, supposed to be carcinogenic in nature. The level of contamination was
15 ppm (by mass).
(i) Express this in
percent by mass.
(ii) Determine the
molality of chloroform in the water sample.
Solution
:
(i) 15 ppm means 5
parts in million (106) parts.
∴ % by mass = 15/106
× 100 = 15 × 10-4 = 1.5×10-3 %
(ii) Molar mass of
chloroform(CHCl3) = 12+1+ (3×35.5) = 118.5 g mol-1
100g of the sample
contain chloroform = 1.5×10-3g
∴ 1000 g (1 kg) of
the sample will contain chloroform = 1.5×10-2 g
= 1.5×10/ 118.65
mole = 1.266 ×10-4 mole
∴ Molality =
1.266×10-4 m.
18. Express the
following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Solution
:-
(i) 0.0048 = 4.8×
10-3
(ii) 234, 000 =
2.34× 105
(iii) 8008 =
8.008× 103
(iv) 500.0 =
5.000× 102
(v) 6.0012 =
6.0012× 100
19. 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
Solution
:
(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5
20. Round up the
following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Solution
:
(i) 34.2
(ii) 10.4
(iii) 0.046
(iv) 2810
1.21. 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
(a) Which law of chemical
combination is obeyed by the above experimental data? Give its statement.
(b) Fill in the blanks in the
following conversions:
(i) 1 km = ...................... mm
= ...................... pm
(ii) 1 mg = ......................
kg = ...................... ng
(iii) 1 mL = ......................
L = ...................... dm3
Solution
:
(a) Fixing the
mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and
80 g in the given four oxides. These masses of dioxygen bears a simple whole
number ratio as 2:4:2:5. Hence, the data given will obey the law of multiple
proportions.
The statement is
as follows two elements always combine in
a fixed mass of other bearing a simple ratio to another to form two or
more chemical compounds.
(b) (i)1 km = 1km× 1000m/ 1km ×100cm /1m/ 10mm /1cm = 106mm
1 km = 1km× 1000m / 1km × 1pm/ 10-12m =
1015 pm
(ii) 1 mg = 1mg ×1g/
1000mg × 1kg / 1000g = 10-6 kg
1 mg = 1mg ×1g/
1000mg × 1ng/ 10-9g = 106 ng
(iii) 1 mL =
1mL×1L/ 1000mL = 10-3 L
1 mL = 1cm3
= 1cm3 × (1dm × 1dm × 1dm/ 10cm × 10cm × 10cm) = 103 dm3
1.22. If the speed of
light is 3.0 × 108ms-1 , calculate the distance covered
by light in 2.00 ns.
Solution
:
Distance covered =
Speed × Time = 3.0 × 108 ms-1
× 2.00 ns
= 3.0 × 108ms-1
× 2.00 ns ×10-9s /1ns =
6.00×10-1m = 0.600m
1.23. 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
:
(i) According to
the reaction, 1 atom of A reacts with 1 molecule of B.
∴200 molecules of B will react with 200 atoms
of A, thereby leaving 100 atoms of A unreacted. Hence, B is the limiting
reagent.
(ii) According to
the reaction, 1 mol of A reacts with 1 mol of B.
∴ 2 mol of A will
react with only 2 mol of B leaving 1 mol of B. Hence, A is the limiting
reagent.
(iii) 1 atom of A
combines with 1 molecule of B.
∴ All 100 atoms of
A will combine with all 100 molecules of B. Hence, the mixture is
stoichiometric and ther is no limiting reagent.
(iv) 1 mol of atom
A combines with 1 mol of molecule B.
∴ 2.5 mol of B
will combine with only 2.5 mol of A. and 2.5 mol of A will be left unreacted.
Hence, B is the limiting reagent.
(v) 1 mol of atom
A combines with 1 mol of molecule B.
∴ 2.5 mol of A
will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be
left. Hence, A is the limiting reagent.
1.24. 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 2.00×103g dinitrogen reacts with 1.00×103g
of dihydrogen.
(ii) Will any of the
two reactants remain unreacted ?
(iii) If yes, which
one and what would be its mass ?
Solution
:
1 mole of
dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of
ammonia (34g).
∴ 2000 g of N2
will react with H2 = 6/28 × 200g = 428.6g. Thus, here N2 is
the limiting reagent while H2 is in excess.
28g of N2
produce 34g of NH3.
∴2000g of N2 will
produce = 34/28 × 2000g = 2428.57 g of NH3.
(ii) N2
is the limiting reagent and H2 is the excess reagent. Hence, H2
will remain unreacted.
(iii) Mass of
dihydrogen left unreacted = 1000g - 428.6g = 571.4 g
1.25. How are 0.50 mol
Na2CO3 and 0.50 MNa2 CO3 different?
Solution
:
Molar mass of
Na2CO3 = (2×23) +12.00+(3×16) = 106 g mol-1
∴0.50 mol Na2CO3
means 0.50 ×106g = 53g
0.50 M Na2CO3
means 0.50 mol of Na2CO3 i.e. 53g of
Na2CO3 are present in 1litre of the solution.
1.26. If ten volumes
of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of
water vapour would be produced ?
Solution
:
Dihydrogen gas
reacts with dioxygen gas as,
2H2(g)
+ O2(g) → 2H2O(g)
Thus, 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.
1.27. Convert the
following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Solution :
(i) 1 pm = 10-12
m
chapter 1-Some
Basic Concepts of Chemistry 28.7 pm = 28.7 × 10-12 m = 2.87 × 10-11
m
(ii) 1 pm = 10-12
m
∴15.15 pm = 15.15
× 10-12 m = 1.515 × 10-11 m
(iii) 1 mg = 10-3
g
25365 mg = 2.5365
× 104×10-3 g
Now,
1 g = 10-3
kg
2.5365×10 g = 2.5365
× 10×10-3 kg
∴25365 mg = 2.5365
× 10-2 kg
1.28. 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)
Solution
:
(i) 1 g Au = 1/197
mol = 1/197 × 6.022×1023 atoms
(ii) 1 g Na = 1/23
mol = 1/23 × 6.022×1023 atoms
(iii) 1 g Li = 1/7
mol = 1/7 × 6.022×1023 atoms
(iv) 1 g Cl2
= 1/71 mol = 1/71 × 6.022×1023 atoms
Thus, 1 g of Li
has the largest number of atoms.
1.29. Calculate the
molarity of a solution of ethanol in water in which the mole fraction of
ethanol is 0.040 (assume the density of water to be one).
Solution
:
Mole fraction of C2H5OH
= No. of moles of C2H5OH /No. of moles of solution
n(C2H5OH)
= n(C2H5OH) / (C2H5OH) + n(H2O)
= 0.040 (Given) ... 1
We have to find
the number of moles of ethanol in 1L of the solution but the solution is
dilute. Therefor, water is approx. 1L.
No. of moles in 1L
of water = 1000g /18g mol-1 =
55.55 moles
Substituting
n(H2O) = 55.55 in equation 1
n(C2H5OH)
/ (C2H5OH) + 55.55 = 0.040
⇒ 0.96n (C2H5OH)
= 55.55 × 0.040
⇒ n(C2H5OH)
= 2.31 mol
Hence, molarity of
the solution = 2.31M
1.30. What will be the
mass of one 12 C atom in g ?
Solution
:
1 mol of 12 C
atoms = 6.022 ×1023 atoms = 12g
∴ Mass of 1 atom
12C = 12 /6.022 ×1023 g = 1.9927× 10-23 g
1.31. How many
significant figures should be present in the answer of the following
calculations?
(i) 0.02856 ×
298.15 × 0.112 /0.5785
(ii) 5 × 5.364
(iii) 0.0125 +
0.7864 + 0.0215
Solution :
(i) Least precise
term i.e. 0.112 is having 3 significant digits.
∴ There will be 3
significant figures in the calculation.
(ii) 5.364 is
having 4 significant figures.
∴ There will be 4
significant figures in the calculation.
(iii) Least number
of decimal places in each term is 4.
∴ There will be 4
significant figures in the calculation.
1.32. Use the data
given in the following table to calculate the molar mass of naturally occurring
argon isotopes:
Isotope Isotopic molar mass Abundance
36Ar 35.96755 g mol-1 0.337%
38Ar 37.96272 g mol-1 0.063%
40Ar 39.9624 g mol-1 99.600%
Solution
:
Molar mass of Ar
= ∑piAi
= (0.00337 × 35.96755 )+ (0.00063 × 37.96272
)+(0.99600 × 39.9624 ) = 39.948 g mol-1
1.33. Calculate the
number of atoms in each of the following
(i) 52 moles of Ar
(ii) 52 u of He (iii) 52 g of He.
Solution :
(i) 1 mol of Ar =
6.022 × 1023 atoms
∴ 52 mol of Ar =
52 × 6.022×1023atoms = 3.131 × 1023 atoms
(ii) 1 atom of He
= 4 u of He
4 u of He = 1 Atom
of He
∴ 52 u of He = 1/4
× 52 = 13 atoms
(iii) 1 mol of He
= 4 g = 6.022 × 1023 atoms
∴ 52 g of He =
(6.022 × 1023/4) × 52 atoms = 7.8286 × 1024 atoms
1.34. A welding fuel
gas contains carbon and hydrogen only. Burning a small sample of it in oxygen
gives 3.38 g carbon dioxide , 0.690 g of water and no other products. A volume
of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate
(i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
Solution
:
Amount of carbon
in 3.38 g of CO2 = 12/44 × 3.38 g = 0.9218 g
Amount of hydrogen
in 0.690 g H2O = 2/18 × 0.690 g = 0.0767 g
The compound
contains only C and H, therefore total mass of the compound = 0.9218 + 0.0767 =
0.9985 g
% of C in the
compound = (0.9218 /0.9985 )×100 = 92.32
% of H in the
compound = (0.0767 /0.9985 )×100 = 7.68
(i) Calculation of
empirical formula,
Moles of carbon in
the compound = 92.32/12 = 7.69
Moles of hydrogen
in the compound = 7.68/1 = 7.68
Simplest molar
ratio = 7.69 : 7.68 = 1(approx)
∴ Empirical
formula CH
(ii) 10.0 L of the
gas at STP weigh = 11.6 g
∴ 22.4 L of the
gas at STP = 11.6/10.0 × 22.4 = 25.984 = 26 (approx)
∴ Molar mass of
gass = 26 g mol-1
(iii) Mass of
empirical formula CH = 12+1 = 13
∴ n = Molecular
Mass/Empirical Formula = 26/13 = 2
∴ Molecular
Formula = C2H2
1.35. Calcium
carbonate reacts with aqueous HCl to give CaCl2 and CO2
according to the reaction, CaCO3 (s) + 2HCl (aq) → CaCl2(aq)
+ CO2 (g) + H2O(l)
What mass of CaCO2
is required to react completely with 25 mL of 0.75 M HCl ?
Solution
:
1000 mL of 0.75 M
HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g
∴ Mass of HCl in
25mL of 0.75 M HCl = 24.375/ 1000 × 25 g = 0.6844 g
From the given
chemical equation,
CaCO3
(s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
2 mol of HCl i.e.
73 g HCl react completely with 1 mol of CaCO3 i.e. 100g
∴ 0.6844 g HCl reacts completely with CaCO3 =
100/73 × 0.6844 g = 0.938 g
1.36. Chlorine is
prepared in the laboratory by treating manganese dioxide (MnO2) with
aqueous hydrochloric acid according to the reaction
4HCl (aq) + MnO2(s)
→ 2H2O(l) + MnCl2(aq) + Cl2(g)
How many grams of HCl
react with 5.0 g of manganese dioxide?
Solution
:
1 mol of MnO2
= 55+32 g = 87 g
87 g of MnO2
react with 4 moles of HCl i.e. 4×36.5 g = 146 g of HCl.
∴ 5.0 g of MnO2 will react with HCl =
146/87×5.0 g = 8.40 g.
Published by lokesh das
