SOME BASIC CONCEPTS OF CHEMISTRY
Type I : Law of Conservation of Mass
⇒
REMEMBER:
“In all physical and chemical changes the total mass of reactants is equal to that of
products.”
1.
What mass of AgNO3 will react with 5.85 g NaCl to produce 14.35 g AgCl and 8.5
g NaNO3, if law of conservation is true?
Ans. AgNO3 + NaCl → AgCl + NaNO3
x
5.85 g
14.35 g
8.5 g
Here,
x + 5.85 = 14.35 + 8.5
x = 17.0 g
If 6.3 g of NaHCO3 are added to 15.0 g of CH3COOH, the residue is found to be
18.0 g. What is the mass of CO2 released?
Ans. CH3COOH + NaHCO3 → CH3COONa + H2O + CO2 ↑
[3.3 gm]
2.
Type II : Law of Constant Composition or Definite proportions
3.
2.16 g of Cu-metal when treated with nitric acid followed by ignition of nitrate
gave 2.70 g of copper-oxide, In another experiment 1.15 g of copper oxide upon
reduction with hydrogen gave 0.92 g Cu. Show that this illustrates law of
definite proportions.
2.16
Ans. i) % of Cu in CuO =
× 100 = 80 %
2.70
∴ Oxygen = 20 %
0.92
ii) % of Cu in 2nd experiment =
× 100 = 80 %
1.15
∴ Oxygen = 20 %
As % is same in both, law is obeyed.
4.
Silver chloride is prepared by,
i) Dissolving 0.5 g of silver wire in HNO3 and adding excess HCl to AgNO3
formed. The weight of this dried AgCl is 0.66 g.
ii) Heating 1 g of silver metal in a current of dry chlorine gas till metal is
completely converted to its chloride. It is found to weigh 1.32 g. Show that
these data obey law of constant composition.
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Type III : Law of Multiple proportions
5.
Two oxides of metal contain 27.6 % and 30.0 % of oxygen respectively. If the
formula of 1st oxide is M3O4, find that of the second.
Ans. 1st oxide Mass of oxygen : 27.6
Mass of metal : 100 - 27.6 = 72.4
nd
2 oxide Mass of oxygen : 30
Mass of metal : 100 - 30 = 70
Formula of 1st oxide : M3O4
3
∴ No. of atoms of metal in 2nd oxide =
× 70 = 2.9
72.4
4
No. of atoms of oxygen in 2nd oxide =
× 30 = 4.35
27.6
∴ Ratio of metal : oxygen in 2nd oxide = 2.9 : 4.35 or 2 : 3
∴ Formula of 2nd oxide is M2O3.
6.
Copper gives two oxides. On heating 7.0 g of each in hydrogen gas 0.888 g and
0.799 g of metal are produced. Show that the results obey law of multiple
proportions.
7.
A metal forms two oxides. One contains 46.67 % of metal and other contains
63.94 % metal. Show that the results are as per law of multiple proportions.
Type IV : Law of Reciprocal proportion
8.
Ammonia contains 82.35 % nitrogen and 17.65 % hydrogen. Water contains 88.9%
oxygen and 11.1 % hydrogen. Nitrogen trioxide contains 63.15 % oxygen and
36.85 % of nitrogen. Show that these data are as per law of reciprocal proportion.
Ans. In NH3,
11.1
17.65
NH
H O
3
2
82.35
1 g of H combines with
82.35
= 4.67 g of nitrogen
17.65
36.85
88.90
NO
2
3
63.15
In H2O,
88.90
= 8.01 g of oxygen
11.1
Ratio of mass of “N” and “O” combining with 1 g “H” = 4.67 : 8.01 or
1 : 1.72
Ratio of “N” and “O” in N2O3 = 36.85 : 63.15 or 1 : 1.71
1 g of H combines with
∴Ratios are same.
9.
CO2 contains 27.27 % carbon, CS2 contains 15.79 % of carbon and SO2 contains 50
% of sulpher. Are these data in agreement with law of reciprocal proportion?
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10.
61.8 g A combines with 80 g B. 30.9 g of A combine with 106.5 g of C. B and C
combine to form compound CB2. Atomic weights of C and B are respectively 35.5
and 6.6. Show that these data are in accordance with law of reciprocal proportion.
Type V : Calculating average atomic mass
11.
Chlorine has two isotopes of a.m.u. 34.97 and 36.97 respectively. The relative
abundances of these isotopes are 0.755 and 0.245 respectively. Find average
atomic mass of chlorine.
Ans. Average atomic mass = (34.97 × 0.755) + (36.97 × 0.245)
= 35.46
12.
Naturally occurring Boron consists of two isotopes whose atomic masses are
10.01 and 11.01 respectively. The atomic mass of natural Boron is 10.81. Calculate
percentage of each isotope in natural boron.
Ans. 20 % and 80 %
Type VI : Calculating mass of single atom or molecule
REMEMBER:
Mass of single atom or molecule =
Molar mass
6.022 × 1023 ( N A )
13.
Calculate mass of i) One atom of Ag
ii) One molecule of CO2.
Molar mass of Ag
108
Ans. i) Mass of 1 atom of Ag =
=
(NA)
6.022 × 1023
= 1.793 × 10-22 g
Molar mass of CO 2
44
ii) Mass of 1 molecule of CO2 =
=
(NA)
6.022 × 10 23
= 7.307 × 10-23 g
Type VII : Calculation of no. of atoms or molecules in a given mass
of substance
14.
REMEMBER:
No. of atoms or molecules = No. of moles × NA
How many atoms or molecules of sulpher are present in 64.0 g sulpher (S8)?
g
Ans. Molar mass of sulpher (S8) = 8 × 32 = 256
mol
64
∴ No. of moles of S8 =
= 0.25
256
∴ No. of molecules = 0.25 × 6.022 × 1023
= 1.50 × 1023 molecules
∴ No. of S-atoms = 1.50 × 1023 × 8 = 1.20 × 1024 atoms
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15.
Calculate no. of molecules present in
i) 34.20 g of cane-sugar (C12H22O11)
ii) One drop of water having mass 0.05 g
Ans. 6.022 × 1022 and 1.67 × 1021 respectively
16.
Calculate no. of atoms of all constituent elements in 53 g Na2CO3.
Ans. Na=6.022 × 1023 ions; C=3.01 × 1023; O=9.033 × 1023
Type VIII : Calculating no. of atoms and molecules in given volume of gas
REMEMBER:
No. of moles of a gas at STP or NTP =
17.
Vol. of gas in litres
22.4
Calculate the mass of
i) 1 × 1023 molecules of methane
ii) 112 cm3 (ml) of hydrogen at STP.
18.
Calculate the volume occupied at STP by,
i) 14 g of Nitrogen
ii) 1.5 moles of CO2
iii) 1021 molecules of Oxygen
Ans. 11.2 L, 33.6 L, 37.2 ml
19.
Calculate no. of moles of each in following
i) 10 g of CaCO3
ii) 1 × 1023 molecules of CO2
Ans. 0.1, 0.166
20.
Calculate the mass of CO2 which contains the same no. of molecules as are in 40
g of Oxygen.
Ans. 55 g
Calculate the mass of Na2CO3 which will have the same no. of molecules as are
in 12.3 g of MgSO4.7H2O.
Ans. 5.3 g
21.
22.
Calculate the volume occupied at STP by,
i) 16.0 g of Oxygen
ii) 1.5 moles of Oxygen
iii) 6.022 × 1023 molecules of CO2
Ans. 11.2 L, 33.60 L, 22.4 L
23.
Calculate the total no. of electrons present in 1.4 g of nitrogen gas.
[Hint: each molecule of N2 has 14 electrons]
Ans. 4.214 × 1023 electrons
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9.7 × 1017 atoms of Fe weigh as much as 1 cc of H2 at STP. What is the atomic
mass of iron?
Ans. 55.9
24.
25.
Chlorophyll contains 2.68 % of magnesium by mass. Calculate the no. of
magnesium atoms in 2.0 g chlorophyll.
Ans. 1.345 × 1021
Type IX : Calculation of Molarity and Normality of solution obtained upon
mixing two or more solutions
REMEMBER:
M V + M 2 V2 + ......
M= 1 1
V1 + V2 + ......
N=
N1V1 + N 2 V2 + ......
V1 + V2 + ......
26.
Calculate molarities and normalities of the solutions obtained on mixing (i) 100
ml 0.2 M H2SO4 + 50 ml 0.1 M HCl
M V + M 2 V2
(0.2 × 100) + (50 × 0.1)
Ans. M = 1 1
=
= 0.167 M
150
V1 + V2
N V + N 2 V2
(0.4 × 100) + (50 × 0.1)
N= 1 1
=
= 0.3 N
V1 + V2
150
[Note: 0.2 M H2SO4 = 0.4 N H2SO4; 0.1 M HCl = 0.1 N HCl]
27.
Calculate molarities and normalities of solutions obtained upon mixing
(i) 100 ml 0.2 N H2SO4 + 50 ml 0.1 N HCl
Type X : Calculations based on Normality and Molarity
REMEMBER:
Mass of solute dissolved in grams
1
Normality =
×
Equivalent mass
Volume of solution in litres
Mass of solute dissolved in grams
1
Molarity =
×
Molecular mass of solute
Volume of solution in litres
Molarity × n = Normality
Calculate molarity of water if its density is 1000 kg/m3.
kg 1000 × 1000 g
g
Ans. d = 1000 3 =
=1
m
1000 × 1000 ml
ml
1 lit. vol. of water has 1000 g
1000
∴No. of moles of water present in 1 lit =
= 55.56 M
18
28.
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29.
A solution of oxalic acid (COOH)2.2H2O is prepared by dissolving 0.63 g of acid
in 250 cm3 of solution. Calculate (i) molarity and (ii) normality.
g
[Mol. mass of (COOH)2.2H2O = 12.6
]
mol
0.63
1
Ans. Molarity =
×
= 0.02 M
126 0.250
Normality = Molarity × n
= 0.02 × 2 = 0.04 N
30.
Commercially available conc. HCl contains 38% HCl by mass. Find
g
molarity.[density=1.19 3 ]
cm
Ans. 38% by mass = 38 g HCl in 100 g solution
100
∴38 g HCl in
ml = 84 ml solution
1.19
38
1
∴ Molarity =
×
= 12.39 M
36.5 0.084
Type XI : If a solution of fixed concentration is diluted then final
normality/molarity can be calculated as
Initial Conc. Final Conc.
=
M1 × V1
M 2 × V2
or
N1 × V1 = N 2 × V2
31.
What volume of 12.38 M HCl is needed to make 1.00 L solution of 0.1 M?
Ans. M1V1 = M2V2
0.1 × 1000
∴ V1 =
= 8.07 ml
12.38
32.
Concentrated H2SO4 is 98% by mass having density of 1.84 g/cc. What volume of
H2SO4 is needed to make 5.0 litre solution of 0.5 M? Find molarity of H2SO4
solution.
Ans. 98% = 98 g H2SO4 in 100 g solution
100
∴ 98 g H2SO4 in
= 54.34 ml solution
1.84
98
1
Molarity =
×
= 18.4 M
98 0.05434
Now, M1V1 = M2V2
0.5 × 5
∴ V1 =
= 0.1358 L = 135.8 ml
18.4
33.
A sample of NaOH weighing 0.40 g is dissolved in water and the solution is
made to 50.0 cm3 in volumetric flask. What is the molarity of resulting solution?
Ans. 0.2 M
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How many moles and how many grams of NaCl are present in 250 cm3 of a 0.50
M NaCl solution?
Ans. 0.125 mole; 7.312 g
34.
35.
In a reaction vessel, 0.184 g NaOH is required to be added for completing the
reaction. How many millilitres of 0.150 M NaOH solution should be added?
Ans. 30.7 ml
Type XII : Finding molality, mass-percent and mole fraction
REMEMBER:
Mass of solute dissolved in grams
1
×
Molality =
Mol. mass of solute
Mass of solvent in kg
Moles of solute
Mole fraction of solute =
Moles of solute + moles of solvent
Mass of solute
Mass-percent =
× 100
Mass of solution
36.
A solution is prepared by adding 2 g of substance “A” to 18 g water. Calculate
mass-percent.
2
× 100 = 10%
Ans. Mass-percent =
[2 + 18]
37.
Calculate mole-fraction of solute in 1 molal aqueous solution.
Ans. 1 molal aqueous solution = 1 mole solute in 1000 g H2O
1
1
∴Mole fraction of solute =
=
= 0.017
1000
[1 +
] 56.5
18
g
38.
The density of 3 M solution of NaCl is 1.25
.Calculate molality of solution.
ml
Ans. 3 molar solution = 3 moles NaCl in 1000 ml solution
For molality, mass of solute and mass of solvent are needed.
From the given data,
Mass of NaCl in solution = 3 × 58.5[mol. mass of NaCl] = 175.5 g
Mass of solution = volume × density = 1000 × 1.25 = 1250 g
∴Mass of solvent = [Mass of solution-Mass of solute]
= [1250-175.5] g
= 1074.5 g
175.5
1
∴ Molality =
= 2.79 m
×
58.5 1.0745
39.
What is the mass percent of solute in the solution obtained by dissolving 5 g of
solute in 50 g of water?
Ans. 9.1 %
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Type XIII :
Calculations involving acid-solution mixed with a base
solution
⇒
REMEMBER:
When acid is neutralized by a base then,
Equivalents of acid = Equivalents of base
i) Finding equivalents in solution of given concentration
Vol. of solution in litres × Normality = Equivalents
ii) Finding equivalents if mass is given in grams
Mass of substance given in grams
Equivalents =
Equivalent mass of substance
40.
Calculate normality and molarity of solution obtained by mixing:
(a) 100 ml 0.1 M H2SO4 + 50 ml 0.1 M NaOH
(b) 50 ml 0.1 N H2SO4 + 100 ml 0.1 N HCl
Ans. (a) Equivalents of 0.1 M H2SO4
0.1 M H2SO4 = 0.2 N H2SO4 [n=2]
∴Equivalents = 0.1 × 0.2 = 0.02
Equivalents of 50 ml 0.1 M NaOH
0.1 M NaOH = 0.1 N NaOH [n=1]
∴Equivalents = 0.05 × 0.1 = 0.005
∴Equivalents of H2SO4 = 0.02 - 0.005
Left unneutralized = 0.015
Equivalents
0.015
∴ Normality of H2SO4 =
=
= 0.1
Vol. of solution in litres 0.150
∴ Normality = 0.1 N
0.1
= 0.05 M
Molarity =
2
(b) Same as [a]
[0.033 N; 0.033 M]
41.
Two acids H2SO4 and H3PO4 are neutralized separately by same amount of alkali
when sulphate and dihydrogen orthophosphate are formed respectively. Find
ratio of mass of H2SO4 and H3PO4.
Ans. 1 : 2
Success seems to be largely a matter of hanging
on after others have let go.
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