NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of
Chemistry
NCERT
Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry are
been solved by expert teachers of CBSETuts.com. All the solutions given in this
page are solved based on CBSE Syllabus and NCERT guidelines.
NCERT Exercises
Question
1.
Calculate the molecular mass of the following :
(i) H2o
(ii) CO2
(iii) CH4
Solution:
(i) Molecular mass
of H2O :2 × 1 + 1 × 16 = 18u
(ii) Molecular
mass of CO2 :1
× 12 + 2× 16 = 44 u
(iii) Molecular
mass of CH4 :
12 + 4 × 1 = 16 u
Question
2.
Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).
Solution:
Molecular mass of Na2SO4 = 2 × Atomic mass of Na + Atomic
mass of S + 4 × Atomic mass of O
= 2 × 23 + 32 + 4 × 16 = 46 + 32 + 64 = 142 u.
Question
3.
Determine the empirical formula of an
oxide of iron which has 69.9% iron and 30.1% oxygen by mass.
Solution:
Hence, the empirical formula is Fe2CO3
Question 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:
(i)
Hence, 1 mole of C produces 44 g of CO2
(ii)
Hence, O2 is the limiting reagent.
∵ 32 g O2 reacts with C to produce 44 g of
CO2
∵ 16 g O2 reacts with C to produce 4432×16=22gofCO2
(iii)
∵ 64 g O2 reacts with C to produce 88 g of
CO2
∵ 16 g O2 reacts with C to produce 8864×16= 22gofCO2
Question
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 82.0245 g mol-1.
Solution:
Question
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 69 means that 69 g of HNO3 are dissolved in 100 g of the solution.
Question 7.
How much copper can be obtained from 100 g of copper sulphate (CuS04) ?
Solution:
Molar mass of CuSO4 =
63.5 + 32 + 4 × 16 =63.5 + 32 + 64 = 159.5 amu or u
159.5 g of CuSO4 contains
copper = 63.5 g
100 g of CuSO4 contains
copper = 63.5159.5×100 = 39.81 g
Question 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.
Solution.
For empirical formula, Fe2O3.
Molecular mass of Fe2O3 x 1 = 2 x 56 + 3 x 16 = 112
+ 48 = 160
Molecular formula = u (Empirical formula)
∴ n=MolecularmassEmpiricalformulamass=160160⇒n=1
∴ Molecular
formula = (Fe2O3) x 1 = (Fe2O3)
Question
9.
Calculate the average atomic mass of chlorine using the following data.
Solution.
Question
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.
Number of moles of carbon atoms in 3 moles of C2H6 = 3 x 2 = 6
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms.
Number of moles of hydrogen atoms in 3 moles of C2H6 = 3 x 6 = 18
(iii) 1 mole of C2H6 = 6.022 x 1023 molecules Number of molecules in 3 moles of C2H6
= 3 x 6.022 x 1023 =
1.807 x 1024 molecules
Question
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 2 L?
Solution.
The molecular mass of sugar ((C12H22O11) = 12 x 12 + 1 x 22 + 11 x 16 = 144 +
22 + 176 = 342
Question
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.
Moles of methanol present in 2.5 L of 0.25 M solution
Question 13.
Pressure is determined as force per unit area of the surface. The SI unit of
pressure, pascal is 1 Pa = 1 Nm-2. If mass of air at sea level is 1034 g cm-2, calculate the pressure in pascal.
Solution.
Mass of air at sea level = 1034 g cm-2
Acceleration due to gravity, g = 9.8 m s-2
Pressureofair=10341000×9.8×100×100=1.01332×105Pa
Question
14.
What is the S.I. unit of mass? How is it defined?
Solution.
The
Question
15.
Match the following prefixes with their
multiplies
Prefixes
Multiples
(i) micro
106
(ii) deca
109
(iii) mega
10-6
(iv) giga
10-15
(v) femto
10
Solution.
(i) micro → 10-6
(ii) deca → 10
(iii) mega → 106
(iv) giga → 109
(v) femto → 10-15
Question
16.
What do you mean by significant figures?
Solution.
Significant figures : The significant figures in a number are all the certain
digits plus one doubtful digit, e.g., 2005 has four significant figures.
Question
17.
A sample of drinking water was found to be severely contaminated with
chloroform, CHCI3,
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
15 parts in million (106)
parts
Question
18.
Express the following in the scientific
notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Solution.
(i) 4.8 x 106
(ii) 2.34 x 105
(iii) 8.008 x 103
(iv) 5.00 x 102
(v) 6.0012 x 100 or 6.0012
Question
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
Question
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) 0.0460
(iv) 2810 or 2.81
x 103
Question
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
16g
(ii) 14 g
32 g
(iii) 28 g
32 g
(iv) 28 g
80 g
(a) Which law of
chemical combination is obeyed bythe 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) Law of multiple proportions : This
law was proposed by
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 are in the ratio of 2 : 4 : 2 :
5 which is a simple whole number ratio. Hence, the given data obeys the law of
multiple proportions.
(b)
(i) 106mm, 1015pm
(ii) 10-6 kg, 106ng
(iii) 10-3 L, 10-3 dm3
Question
22.
If the speed of light is 3.0 x 108 m s-1,
calculate the distance covered by light in 2.00 ns.
Solution.
Speed of light = 3.0 x 108 m
s-1
Distance covered by light in 2.00 ns = 3.0 x 108 x 2 x 10-9 = 6.00 x 10-1 m = 0.600 m
Question
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.
According to the equation, one mole of A reacts with one mole of B and one atom
of A reacts with one molecule of B.
(i) B is limiting
reagent because 200 molecules of B will react with 200 atoms of A and 100 atoms
of A will be left in excess.
(ii) A
(iii) Both will react
completely because it is stoichiometric mixture. No limiting reagent.
(iv) B
(v) A
Question
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 x 103 g dinitrogen reacts with 1.00 x 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.
The balanced chemical equation is N3 + 3H3 → 2NH3
Question 25.
How are 0.50 mol Na2CO3 and 0.50 M Na2CO3different?
Solution.
1 mol Na2CO3 ≡ 2 x 23 + 12 + 3 x 16 = 106 g mol-1
0.50 mol Na2CO3 ≡ 0.50
x 106= 53 g
0.50 M Na2CO3 solution means that 0.50 moles
or 53 g of Na2CO3 are dissolved in 1000 mL of
solution.
Question
26.
lf ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how
many volumes of water vapour would be produced?
Solution.
Question
27.
Convert the following into basic units :
(i) 28.7 pm
(ii) 15.15pm
(iii) 25365 mg
Solution.
Question
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.
Question
29.
Calculate the molarity of a solution of ethanol in water in which the mole
fraction of ethanol is 0.040.
Solution.
Question
30.
What will be the mass of one 12C atom in g?
Solution.
Question 31.
How many significant figures should be present in the answer of the following
calculations?
(i) 0.02856×298.15×0.1120.5785
(ii) 5×5.364
(iii) 0.0125 +
0.7864 + 0.0215
Solution.
Question
32.
Use the data given in the following table to calculate the molar mass of
naturally occurring argon isotopes:
Solution.
Question
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.
Question
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.
Question
35.
Calcium carbonate reacts with aqueous HCI to give CaCI2 and CO2 according to the reaction,
CaC03(s) + 2HCI(aq) —> CaCI2(aq)+ CO2(g) + H2(s)O(I).
What mass of CaC03 is required to react completely with 25 mL of 0.75 M HCI?
Solution.
Question
36.
Chlorine is prepared in the laboratory by treating manganese dioxide (Mn02)
with aqueous hydrochloric acid according to the reaction
4HCI(aq) + MnO2(s) → 2H2O(I) + MnCI2(aq) + Cl2(g)
How many grams of HCI react with 5.0 g of manganese dioxide?
Solution.
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