Some Basic Concepts of Chemistry - 1 File Download

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Level - I

Chapter 1

Some Basic Concepts of Chemistry

SECTION - A

School/Board Exam. Type Questions

Very Short Answer Type Questions:

1. What is the SI unit of volume? What is the other common unit which is not a SI unit of volume?

Sol. The SI unit of volume is m3 whereas litre (L) is the common unit which is not an SI unit.

2. What are the reference points in thermometer with celsius scale?

Sol. The thermometers with celsius scale are callibrated from 0° to 100° where there two temperatures are thefreezing and boiling of water.

3. What is the SI unit of density?

Sol. The SI unit of density is kgm–3 or kg/m3.

4. What information we get from a balanced chemical equation?

Sol. A chemical equation in which the number of atoms of each element is equal on the reactant side and theproduct side is called a balanced equation.

5. Define molarity of a solution. Give its SI units.

Sol. Molarity is defined as the number of moles of solute per litre of the solution. Its SI unit is mol L–1.

6. What is the effect of temperature on followings?

(a) Molarity (b) Mole fraction

Sol. (a) Molarity decreases with increase in temperature

(b) Mole fraction : It remains unchanged with change in temperature.

7. Why is the law of Gay-Lussac not obeyed if any reactant or product is not a gas?

Sol. If any reactant or product is a liquid, the volume occupied by a liquid is extremely small and hence, the lawis not obeyed.

8. Calculate the mass of 3 g molecules of HNO3.

Sol. 1 g molecule of HNO3 = 1 + 14 + (3 × 16) = 63 g

3 g molecule of HNO3 = 189 g

Solutions (Set-1)

2 Some Basic Concepts of Chemistry Solutions of Assignment (Set-1) (Level-I)


9. How many significant figures are there in each of the following numbers?

(a) 15

(b) 1.053

(c) 0.0050

Sol. (a) Two (b) Four (c) Two

10. Perform the following calculations to the proper number of significant digits

(a) (2.00 × 10–2 km) + (4.2 × 10–2 km)

(b) (1.5 × 101 cm) × (8.0 × 102 cm) × (0.0100 m)

Sol. (a) 6.2 × 10–2 km

(b) (1.5 × 101 cm) × (8.0 × 102 cm) × (1 cm) = 12 × 103 cm3

1.2 × 104 cm3

Short Answer Type Questions:

11. What does the following prefixes stand for?

(a) pico (b) nano

(c) centi (d) deci

Sol. (a) pico = 10–12 (b) nano = 10–9

(c) centi = 10–2 (d) deci = 10–1

12. Convert 35°C to °F and K.

Sol.9

F ( C) 325

9(35) 32

5 = 63 + 32 = 95°F

K = °C + 273.15 = 35 + 273.15 = 308.15 K

13. (i) Calculate the percentage composition of the named element in the following compounds

(a) Al in KAl(SO4)2 12H

2O

(b) Na in Na2SO

4 10H

2O

(ii) How many moles are present in following?

(a) 24.5 g H2SO

4

(b) 4.00 g O2

Sol. (i) (a) Molar mass of KAl(SO4)2 12H

2O

= 39 + 27 + (2 × 96) + (12 × 18)

= 474 g mol–1

Percent by mass of Al in KAl(SO4)2 12H

2O

27100 5.69%

474

(b) Molar mass of Na2SO

4 10H

2O

= 2 × 23 + 96 + 10 × 18

= 322 g mol–1

Percent by mass of Na 23100 7.14%

322

3Solutions of Assignment (Set-1) (Level-I) Some Basic Concepts of Chemistry


(ii) (a) 98 g H2SO

4 = 1 mol

24.5 g H2SO

4

24.50.25 mol

98

(b) 32 g of O2 = 1 mol

4.00 g of O2 1

0.125 mol8

14. State the rules to follow in rounding off a number.

Sol. (i) If the digit dropped is greater than 5, add 1 to the last remaining digit.

e.g., 62.138 will become 62.14

(ii) If the digit dropped is less than 5, the last remaining digit is not changed.


(iii) If the digit dropped is 5, the last remaining digit is left unchanged if it is even; 1 is added if it is odd.


1.8235 will become 1.824

15. If the speed of Shatabdi express is 150 miles per hour, then express its speed in SI unit.

(1 mile = 1.6 km)

Sol. The SI unit for speed are ms–1

As given, 1 mile = 1.6 km = 1.60 × 103 m

Conversion factor 3

1.60 10 m

1 mile

1 hour = 60 × 60 s = 3.6 × 103 s

Conversion factor 3

3.6 10 s

1 hour

Now speed 150 miles

hour

3

3

150 miles 1.60 10 m 1 hr

hour 1 mile 3.6 10 s

= 66.66 ms–1

16. What is the law of constant proportions? Write with an example.

Sol. Law of constant proportions was proposed by Joseph Proust. This law states that “a chemical compound always

contains same elements combined together in same proportion by mass”.

For example, pure water obtained from different sources such as, river, well, spring, sea, etc., always contains

hydrogen and oxygen combined together in the ratio 1 : 8 by mass.

17. Copper and oxygen combine to form two oxides, the red cuprous oxide (Cu2O) and the black cupric oxide

(CuO). Show that these compounds follow the law of multiple proportions.

Sol. In red oxide (Cu2O) : 16 parts by mass of oxygen combine with 63.5 × 2 parts by mass of copper.

In black oxide (CuO) : 16 parts by mass of oxygen combine with 63.5 parts by mass of copper.

Ratio of masses of copper that combine with fixed mass of oxygen (16 parts). In these oxides is 63.5 × 2 : 63.5

or 2 : 1, which is a simple whole number ratio.



18. Calculate the mole fraction of ethyl alcohol (C2H5OH) and water in a solution in which 46 g of ethyl alcohol

and 90 g of water have been mixed.

Sol.2 5

C H OHw 46 g

Meth

= 46 g mol–1

2 5

C H OH

46n 1 mol

46

2

2

2

H O

H O

H O

w 90n 5 mol

M 18

ntotal

= 6 mol

2 5

2

eth

C H OH

eth H O

n 1x 0.167

n n 6

and2

H O

5x 0.833

6

19. A battery acid contains 24.5% by mass of H2SO

4. What is the molality of the solution?

Sol. 24.5% by mass H2SO

4 means 100 g solution contain 24.5 g H

2SO

4

2 4

H SOw 24.5 g

2 4H SO

24.5n 0.25 mol

98

2H O

w 75.5 g

B

A

n 0.25Molality (m) 1000 1000

w 75.5

m = 3.31 m kg–1

20. A sample of sodium hydroxide weighing 0.48 g is dissolved in water and the solution is made to50.0 cm3 in a volumetric flask. What is the molarity of the resulting solution?

(molar mass of NaOH = 40)

Sol. Mass in g of NaOH = 0.48 g

Number of moles of NaOH 20.48

1.2 10 mol40

Volume of solution = 50 cm3 = 0.05 L

2

1

2

1.2 10 molMolarity 0.24 M = 0.24 mol L

5.0 10 L

21. What is the molality of a 1 L solution of H2SO

4 which is 46% wt/vol?

(Density of solution = 1.40 g/cm3)

Sol. 46 wt/volume solution means 100 mL solution contains 46 g H2SO

4. So 1 L (1000 mL) solution contains 460 g

H2SO

4.

Therefore, number of moles of H2SO

4

4604.69

98

and mass of solution = vol. × density = 1400 g



Then, mass of the solvent = mass of solution – mass of solute

= 1400 – 460 = 940 g

We know that

solute

solvent

n 4.69Molality(m) 1000 1000

w 940

= 4.98 mol kg–1

22. An aqueous sugar syrup of mass 224.2 g contains 34.2 g of sugar (C12

H22

O11

). Calculate

(i) the molality of the solution.

(ii) the mole fraction of sugar solution.

Sol. Molar mass of sugar = 342 g/mol

Moles of sugar 34.20.1

342

Mass of water = mass of solution – mass of sugar

= 224.2 – 34.2 = 190 g

(i) moles of soluteMolality 1000

mass of solvent (in grams)

–10.11000 0.52 mol kg

190

(ii) Mole fraction of sugar

2

sugar 3

sugar H O

n 0.1 0.19.38 10

190n n 10.650.1

18

23. Calculate the molarity of pure water if its density at room temperature is 1.0 g/cm3.

Sol. Let the volume of water be 1 L = 1000 cm3

Mass of water = volume × density

= 1000 × 1 = 1000 g

Moles of water 1000

55.5518

and molarity of water 55.55

55.55 mol/L1

24. The following reaction,

2Br–(aq) + Cl2(aq) 2Cl–(aq) + Br

2(aq)

is used for the commercial preparation of bromine from its salts. Suppose we have 50.0 mL of a 0.080 M

solution of NaBr. What volume of a 0.050 M solution of Cl2 is needed to react completely with Br–?

Sol. The reaction given is

2Br–(aq) + Cl2(aq) 2Cl–(aq) + Br

2(aq)

Number of moles of Br– = Number of moles of NaBr

1 50.00.080 mol L L

1000



= 4.0 × 10–3

Number of moles of Cl2 required

= 1

2 × No. of moles of Br– = 2 × 10–3 mol

The volume of aqueous Cl2 needed

2Number of moles of Cl

Molarity

3

1

2 10 mol0.04 L = 40 mL

0.050 mol L

25. Write four points of Dalton’s atomic theory.

Sol. In 1808, Dalton published “A new system of chemical philosophy” in which he proposed the following :

1. Matter consists of indivisible atoms

2. All the atoms of a given element have identical properties including identical mass. Atoms of different

elements differ in mass.

3. Compounds are formed when atoms of different elements combine in a fixed ratio.

4. Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical

reaction.

26. Calculate the number of atoms in each of the following

(i) 4 mole atoms of nitrogen.

(ii) 0.4 mole molecules of nitrogen.

Sol. (i) 1 mole of atom of nitrogen = 6.022 × 1023 atoms

4 mole atoms of nitrogen = 4 × 6.022 × 1023 = 24.08 × 1023 atoms

(ii) 1 mole molecule of nitrogen = 6.022 × 1023 molecules

0.4 mole molecules of nitrogen

= 0.4 × 6.022 × 1023

= 2.408 × 1023 molecules

= 2 × 2.408 × 1023 atoms

= 4.816 × 1023 atoms

27. Express the following in the scientific notation

(i) 1,66,000

(ii) 1660

(iii) 0.0016

(iv) 166.0

Sol. (i) 1.66 × 105

(ii) 1.660 × 103

(iii) 1.6 × 10–3

(iv) 1.6 × 102



28. Express the following SI base units using power of 10 notations

(i) 2.35 millimoles

(ii) 2 day

(iii) 8.45 mL

(iv) 68 g

(v) 0.0826 inch

Sol. (i) 1 mol = 103 millimole

2.35 millimole = 2.35 × 10–3 mole

(ii) 1 day has 24 hr i.e., 24 × 60 min i.e.,

24 × 60 × 60 s

Therefore, 2 day = 2 × 86400 s = 172800 = 1.72 × 105 s

(iii) 1 m3 = 103 L, therefore 1 mL = 10–6 m3

Hence, 8.45 mL = 8.45 × 10–6 m3

(iv) 1 = 10–6 g

68 g = 68 × 10–6 g = 6.8 × 10–5 g

(v) 1 inch = 2.54 × 10–2 m

0.0826 inches = 0.0826 × 2.54 × 10–2 m

= 2.09 × 10–3 m

29. Zinc and hydrochloric acid react according to the reaction

Zn(s) + 2HCl(aq) ZnCl2(aq) + H

2(g)

If 0.40 mol Zn are added to hydrochloric acid containing 0.62 mol HCl, how many moles of H2 are produced?

Sol. 2 20.40 mol 0.62 mol

Zn(s) 2HCl(aq) ZnCl (aq) H (g)

0.40 mol Zn requires 2 × 0.40 = 0.80 mol HCl but only 0.62 mol of HCl are given.

Therefore, HCl is a limiting reactant.

Number of mole of H2 produced

HCl

1n

2

10.62 0.31 mol

2

30. Calculate the molarity of solution obtained by dissolving 0.100 g of Na2CO

3 in 250 cm3 of solution.

Sol. Molar mass of Na2CO

3 = 106 g mol–1

Number of moles of Na2CO

3 40.100 g

9.4 10 mol106

Volume of solution = 250 cm3 1

L4

4n 9.4 10

M 0.00376 M1v

L4



Long Answer Type Questions

31. Express each of the following in SI units

(i) 83 million mile

(ii) 7 feet 2 inch

(iii) 0.64 Å

(iv) 250 miles per hour

(v) 24 pounds per square inch

(vi) 26°C

(vii) 250 pounds

Sol. (i) 1 mile = 1.6 × 103 m, 1 million = 106

83 million miles = 83 × 106 × 1.60 × 103 m

= 132.8 × 109 = 1.32 × 1011 m

(ii) 1 feet = 12 inch

7 × 12 + 2 = 86 inch

and 1 metre = 39.37 inch

8686 inch = 2.18 m

39.37

(iii) 1 Å = 10–10 m

0.64 Å = 0.64 × 10–10 m

or 6.4 × 10–11 m

(iv) 1 mile = 1.6 × 103 m and 1 hour = 60 × 60 sec

3 3

1

3

250 1.6 10 400 10250 miles/hr 111.11 ms

60 60 3.6 10

(v) 14.7 lb/m2 = 1 atm = 1.013 × 105 pascals

24 pounds per square inch

= 24 × 6894.76 Nm–2

= 165474.24 Nm–2

= 1.65 × 105 pascals

(vi) 26°C = 26 + 273 = 299 K

(vii) 1 lb = 0.45359 kg

250 pound = 250 × 0.45359 = 113.39 kg

32. 6.055 × 10–2 kg of washing soda (Na2CO

3 10H

2O) is dissolved in water to obtain 1 L of a solution of density

1077.2 kg/m3. Calculate the molarity, molality and mole fraction of Na2CO

3 in the solution.

Sol. Molar mass of washing soda is

(2 × 23) + 12 + (3 × 16) + 18 × 10 = 286 g mol–1

or 286 × 10–3 kg mol–1

3

solute 3 1

6.055 10 kgn 0.211 mol

286 10 kg mol



(i) soluten 0.211

Molarity 0.211 Mv 1

Mass of 1 L solution (10–3 m3) = density × volume

= 1077.2 kg m–3 × 10–3 m3

= 1.0772 kg

Mass of solvent = (1.0772 – 0.06055) kg = 1.01665 kg

(ii) 1moles of solute 0.211Molality 0.207 mol kg

mass of solvent in kg 1.01665

(iii)solute

solute

solute solvent

n 0.211x

1016.65n n0.211

18

0.2110.00372

0.211 56.48

33. Calculate the molality of solution of C2H5OH in water in which the mole fraction of C

2H5OH is 0.050.

Sol.2 5

2 5

2 5 2

C H OH

C H OH

C H OH H O

nx 0.050

n n

2H O

n is 1 L H2O

100055.55

18

2 5

2 5

C H OH

C H OH

n0.050

n 55.55

2 5 2 5

C H OH C H OHn 0.050 n 2.77

2 5C H OH

2.77 0.95 n

2 5C H OH

2.77n 2.91 mol

0.95

Now, 1 L H2O = 1 kg H

2O (At 4°C)

2 5

2

C H OH 1

H O

n 2.91Molality 2.91 mol kg

w (in kg) 1

34. A sample of phosphorus has 0.5 moles of P4 molecules

(i) How many P4 molecules are there?

(ii) How many P atoms are there?

(iii) How many moles of P atoms are there in the sample?

(iv) What is the mass of the sample?

Sol. (i) 1 mole of P4 contains 6.022 × 1023 P

4 molecules

Number of P4 molecules in 0.5 mol

236.022 10 molecules

0.5 mol1 mol

= 3.011 × 1023 P4 molecules



(ii) Since, 1 P4 molecule contains 4 P atoms

Number of P atoms = 4 × 3.011 × 1023 = 12.044 × 1023 P atoms

(iii) Moles of P atoms = 0.5 × 4 = 2 mole

(iv) Molecular mass of P4 = 4 × 31 = 124 u

1 mole of P4 molecule weighs 124 grams

mass of 0.5 mol P4 molecules

124 g0.5 mol = 62.0 g

1 mol

35. Calculate the percentage composition of zinc, phosphorus and oxygen in zinc phosphate, Zn3(PO

4)2.

(Zn = 65.5, P = 31 and O = 16 u)

Sol. 1 mole of Zn3(PO

4)2 contains 3 mol of zinc

2 mol of phosphorus and 8 mol of oxygen

Mass of 3 mol of Zn = 3 × 65.5 = 196.5 g

Mass of 2 mol of P = 2 × 31 = 62 g

Mass of 8 mol of O = 8 × 16 = 128 g

Mass of 1 mol of Zn3(PO

4)2 = 196.5 + 62 + 128 = 386.5 g

(a) Percentage composition of Zn

Mass of Zn100

Mass of sample

196.5 g100 50.84% Zn

386.5 g

(b) Percentage composition of P

Mass of P100

Mass of sample

62 g100 16.04% P

386.5 g

(c) Percentage composition of O

Mass of O100

Mass of sample

128 g100 33.12% O

386.5 g

36. 0.90 g of an organic compound containing only carbon, hydrogen and nitrogen on combustion gave 1.1 g CO2

and 0.3 g water. What is %C, %H and %N in the organic compound?

Sol. Mass of organic compound, w = 0.90 g

Mass of CO2, w

1 = 1.1 g

Mass of water, H2O, w

2 = 0.3 g

1 mol C 1 mol CO2

12 g of C = 44 g CO2



(a) Percentage of carbon 2Mass of CO12 g

10044 g Mass of sample

12 1.1 g 1320100 33.33% C

44 0.90 g 39.6

(b) 2 mol H = 1 mol H2O

2 g H = 18 H2O

Percentage of hydrogen 2Mass of H O2 g

10018 g Mass of sample

2 0.3 g 60100 3.70% H

18 0.90 g 16.2

(c) Percentage of nitrogen = 100 – (%C + %H)

= 100 – (33.33 + 3.70)

= 100 – 37.03 = 62.97% N

37. Calculate the number of atoms of oxygen present in 132 g CO2. What would be the weight CO having the

same number of oxygen atom?

Sol. ∵ Molecular mass of CO2 = 44 g/mol

1 molecule of CO2 has 2 oxygen atoms

44 g of CO2 = 2 × 6.022 × 1023 oxygen atom

132 g of CO2 =

23 132 g2 6.022 10

44 mol

= 36.132 × 1023 atoms of oxygen

Molecular mass of CO = 28 g/mol. It has 16 g O and one atom of O in one molecule of CO.

∵ In carbon monoxide, mole of CO = mole of C atom

= mole of O atom

6.022 × 1023 O atoms = 1 CO = 28 g

36.132 × 1023 atoms of O 23

23

28 g/mol 36.132 10 atom168 g CO

6.022 10 atom/mol

38. What is the molecular formula of a compound that contains 47.4% S and 52.6% Cl? The molecular mass ofthe compound as determined experimentally is 135 g mol–1.

Sol.

S

Cl

47.4

52.6

Element Percentage Relative no.of atoms

47.4

32= 1.48 1

1

Simpleratio

Divided bylowest number

32

35.5

Atomicmass

52.6

35.5= 1.48

1.48

1.48= 1

1.48

1.48= 1

The empirical formula of the compound is SCl

Calculation of molecular formula

Empirical formula mass = 1 × 32 + 1 × 35.5 = 67.35 amu

Molecular mass 135n 2

Empirical formula mass 67.5

Molecular formula = n × empirical formula

2 × SCl = S2Cl2



39. A 0.2076 g sample of an oxide of cobalt on analysis was found to contain 0.1476 g cobalt. Calculate the

empirical formula of the oxide. [Co = 59 amu]

Sol. (a) Mass of Co + mass of O = mass of cobalt oxide

Mass of O = mass of cobalt oxide – mass of Co

= 0.2076 g – 0.1476 g = 0.06 g

(b) Moles of Co in the oxide Mass of CO

Atomic mass of cobalt

0.1476 g0.002501 mol

59 g/mol

Moles of O in the oxide Mass of O

Atomic mass of oxygen

0.06 g0.00375 mol

16 g/mol

(c) Mole ratio Co 0.002501 2

0.67O 0.00375 3

(d) Since, the mole ratio of atoms in a sample of a compound is equal to the ratio of atoms in its formula,

the empirical formula of the oxide of cobalt is Co2O3.

40. The mass percent composition of a substance is 24.7% Ca, 1.24% H, 14.8% C and 59.3% O. Calculate the

molecular formula of the substance if its molecular mass is 162 u.

Sol.

Ca

H

C

O

24.7

1.24

14.8

59.3


24.7

40= 0.62 1

2

2

6

Simpleratio

Divided bylowest number

40

1

12

16

Atomicmass

1.24

1= 1.24

0.62

0.62= 1

1.24

0.62= 2

14.8

12= 1.23

59.3

16= 3.71

1.23

0.62= 2

3.71

0.62= 5.98

The empirical formula of the compound is CaH2C2O6

Calculation of molecular formula

Empirical formula mass = 40 + 2 × 1 + 2 × 12 + 6 × 16 = 162


Empirical formula mass 162


= 1 × CaH2C2O6

= CaH2C2O6 or Ca(HCO

3)2



41. The oxides of lead were separately reduced to metallic lead by heating in a current of hydrogen and the followingdata were obtained

(i) Mass of yellow oxide = 3.45 g

Loss of mass during reduction = 0.24 g

(ii) Mass of brown oxide = 1.195 g

Loss of mass during reduction = 0.156 g

Show that the above data illustrate the law of multiple proportions.

Sol. Hydrogen combines with oxygen of the oxide to give metal. Therefore, loss in mass during reduction of theoxide is equal to the mass of oxygen.

(i) Let the fix the mass of oxygen with which metal combines = 1 g

Mass of oxide = 3.45 g

Mass of oxygen = 0.24 g

Mass of lead = 3.45 g – 0.24 g = 3.21 g

0.24 g oxygen combines with 3.21 g lead

1 g oxygen combines with 3.21

0.24 = 13.4 g lead

(ii) Mass of oxide = 1.195 g

Mass of oxygen = 0.156 g

Mass of lead = 1.195 g – 0.156 g = 1.039 g

0.156 g oxygen combines with 1.039 g lead

1 g oxygen combines with 1.039

0.156 = 6.7 g lead

The ratio of the masses of lead that combines with a fixed mass of oxygen (1 g) is 6.7 : 13.4 or 1 : 2.This ratio is simple and therefore illustrates the law of multiple proportions.

42. What will be the concentration of a solution which is made by mixing 25 mL of 0.2 M H2SO

4, 35 mL of 0.25

M H2SO

4 and 45 mL of 0.35 M H

2SO

4?

Sol. Moles of H2SO

4 in 25 mL of 0.2 M H

2SO

4

= M × V in litres

250.2 M × litres = 0.005 mol

1000

Moles of H2SO

4 in 35 mL of 0.25 M H

2SO

4

350.25 M × litres = 0.00875 mol

1000

Moles of H2SO

4 in 45 mL of 0.35 M H

2SO

4

450.35 M × litres = 0.01575 mol

1000

Total moles of H2SO

4 = 0.005 + 0.00875 + 0.01575 = 0.0295 mol

Total volume = 25 mL + 35 mL + 45 mL = 105 mL = 0.105 litre

Final concentration 2 40.0295 mole H SO

0.105 litre of solution = 0.2809 M H

2SO

4



43. A compound of carbon, hydrogen and nitrogen contains these elements in the ratio 9 : 1 : 3.5. Calculate theempirical formula, if its molecular mass is 108, what is the molecular formula?

Sol.

Carbon

Hydrogen

Nitrogen

9

1

3.5

Element Elementratio

Relative no.of atoms

912

= 0.75

Simplestratio

12

1

14

Atomicmass

11

= 1

0.750.25

= 3

10.25

= 4

3.514

= 0.250.250.25

= 1

Empirical formula = C3H4N

Empirical formula mass = (3 × 12) + (4 × 1) + 14 = 54



Thus, molecular formula of the compound = 2 × empirical formula

= 2 × C3H4N

= C6H8N2

44. A carbon compound containing only carbon and oxygen has an approximate molecular mass of 290. On analysisit is found to contain 50% by mass of each element. What is the molecular formula of the compound?

Sol.

Carbon

Oxygen

50.0

50.0


4

3

Simplest wholenumber ratio

12

16

Atomicmass

4.1663.125

= 1.334.166

3.1253.1253.125

= 1

Simplestratio

The empirical formula C4O3

Empirical formula mass = (4 × 12) + (3 × 16) = 96

Molecular mass = 290

Molecular mass 290n 3 approximately

Empirical mass 96


= 3 × C4O3

= C12

O9

45. A compound has following composition

(i) Sodium = 14.13%

(ii) Sulphur = 9.97%

(iii) Oxygen = 69.50%

(iv) Hydrogen = 6.22%.

Calculate the molecular formula of the compound assuming that whole hydrogen in the compound is presentas water of crystallisation. Molecular mass of the compound is 322.



Sol.

Sodium

Sulphur

Hydrogen

Oxygen

14.31

9.97

6.22

69.50


23

32

1

16

Atomicmass

0.6220.311

= 20.622

0.311

6.22

4.34

0.3110.311

= 1

Simplestratio

6.220.311

= 20

4.340.311

= 14

The empirical formula = Na2SH

20O14

Empirical formula mass = (2 × 23) + 32 + (20 × 1) + (14 × 16) = 322

Molecular mass = 322

Molecular formula = Na2SH

20O14

All of the hydrogen is present in the form of water.

Thus, 10 water molecules are present in the molecule.

So, molecular formula = Na2SO

4 10H

2O

SECTION - B

Model Test Paper

Very Short Answer Type Questions :

1. What is the difference between precision and accuracy?

Sol. Precision means the closeness of various measurements for the same quantity. Accuracy is the agreementof a particular value to the true value of the result.

2. What is formula mass?

Sol. When a substance does not contain discrete molecules of their constituent units and have a three dimensionalstructure, formula mass is used to calculate molecular mass which is the sum of the atomic masses of allatoms present in the formula.

3. How many molecules present in one mole of a substance?

Sol. 1 mol = 6.022 × 1023 molecules

4. What is one molal solution?

Sol. One molal solution is that solution which contain 1 mole solute in 1 kg of solvent.

5. Calculate the number of molecules present in 0.5 mole of CO2?

Sol. The number of molecules present in 0.5 moles of CO2 is 6.022 × 1023 × 0.5 = 3.011 × 1023.

6. Write an expression for molarity and molality of a solution.

Sol.Number of moles of solute

MolarityVolume of solution in litres

Number of moles of soluteMolality

Mass of solvent in kg



7. Express 1947, 0.00019, 0.02601, 2600.00 in the scientific notation.

Sol. 1947 = 1.947 × 103

0.00019 = 1.9 × 10–4

0.02601 = 2.601 × 10–2

2600.00 = 2.60 × 103

8. Write two differences between pure substances and mixtures.

Sol. Pure substance have fixed composition and cannot be separated by simple physical methods whereas mixtures

do not have fixed composition or ratio and can be separated by physical methods.

Short Answer Type Questions :

9. Define Kelvin.

Sol. The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of

the triple point of water.

10. Calculate the number of moles in the following masses

(i) 9 g of Fe (ii) 9 mg of Ca

Sol. (i) 7.85 g of Fe

56 g of Fe contains 6.022 × 1023 atoms = 1 mol

56 g of Fe = 1 mol

9 g of Fe 1

956

= 0.16 mol

(ii) Mole of Ca –3

9 10

40

= 2.25 × 10–4

11. Write the empirical formula of the compound having molecular formulae

(i) C6H6

(ii) C6H12

(iii) H2O2

(iv) H2O

Sol. Empirical formula is a simplest whole number ratio of atoms in the molecule, therefore the empirical formula

of given compounds are

(i) CH (ii) CH2

(iii) HO (iv) H2O

12. A solution is prepared by adding 4 g of a substance A to 18 g of water. Calculate the mass percent of the

solute.

Sol. Mass percent of A Mass of A

100Mass of solution

= 4 g

1004 g of A + 18 g of water

4100 18.18%

22



13. How many significant figures should be present in the answer of the following calculations?

(i)3.24 0.08666

5.006

(ii) 0.58 + 324.65

(iii) 1.78986 × 103 (iv) 943 × 0.00345 + 101

Sol. (i) The result of multiplication

3.24 0.08666

5.006

= 0.0560883

0.0561

As the least number of the significant figures is three, therefore the result is rounded off to 0.0561. The 0

is rounded off to 1 because the next digit is greater than 5.

(ii) 5 significant figures

(Answer should be given by least decimal place)

(iii) 6 significant figures

(iv) 3 significant figures

Short Answer Type Questions :

14. Three litres of water are added to 2 L of 5 molar HCl solution. What is the molarity of resulting solution?

Sol. Initial volume, V1 = 2 L

Final volume, V2 = 3 L + 2 L = 5 L

Initial molarity, M1 = 5 M

Final molarity = M2

M1V1 = M

2V2

5 M × 2 L = M2 × 5 L

2

5 M × 2 LM 2 M

5 L

Thus, the resulting solution is 2 M HCl

15. State law of multiple proportions with an example.

Sol. If two elements can combine to form more than one compounds, the masses of one element that combine

with a fixed mass of the other element, are in the ratio of small whole numbers.

For example, hydrogen combines with oxygen to form two compounds, namely water and hydrogen peroxide.



18 g2 g 16 g

Hydrogen Oxygen Water

2 g 32 g 34 g

Hydrogen Oxygen Hydrogen peroxide

Here, the masses of oxygen (i.e., 16 g and 32 g) which combine with the fixed mass of hydrogen (2 g) bear

a simple ratio, i.e., 16 : 32 or 1 : 2.

16. A solution is prepared by adding 60 g CH3OH to 120 g of water. Calculate the mole fraction of CH

3OH and

water.

Sol. Number of moles of solute A

33

(n ) 3

Weight of CH OH(CH OH)

Molecular mass of CH OH

601.875

32

Number of moles of B

(n )

120water 6.667

18

nA + n

B = 1.875 + 6.667 = 8.542

B

1.8750.220

8.542

A = 1 –

B = 0.780

17. Calculate the number of molecules of sulphur (S8) present in 32 g of solid sulphur.

Sol. Given, w = 32 gm

Atomic mass of S = 32

Molar mass of S8 = 32 × 8 = 256

No. of molecules N = no. of moles × 6.022 × 1023

No. of moles w

M 32

0.125256

So, number of molecules = 0.125 × 6.022 × 1023

= 7.52 × 1022 molecules

18. Calculate the molar mass of C2H6, C

12H22

O11

, H2SO

4, H

3PO

4.

Sol. C2H6 = (2 × 12) + (6 × 1) = 30

C12

H22

O11

= (12 × 12) + (22 × 1) + (11 × 16) = 342

H2SO

4 = (2 × 1) + 32 + (4 × 16) = 98

H3PO

4 = (1 × 3) + 31 + (4 × 16) = 98



Long Answer Type Questions :

19. (i) Calculate the mass of 2.5 gram atom of calcium. (Atomic mass of calcium is 40.)

(ii) Calculate the mass of 1.5 gram molecule of water (H2O).

Sol. (i) 1 gram atom of calcium

= gram atomic mass of calcium = 40 g

2.5 gram – atom of calcium

= 40 × 2.5 = 100 g

(ii) Molecular mass of water (H2O)

1 × 2 + 16 = 18 u

1 gram molecule of H2O

= gram molecular mass of H2O = 18 g

1.5 gram molecule of H2O = 1.8 × 1.5 = 27 g

20. Calculate number of atoms of each type in 6.84 g of sucrose (C12

H22

O11

).

OR

An organic compound contains element C, H and oxygen. A 4.24 mg sample of compound is completely burnt

in oxygen. It gives 8.45 mg of carbon dioxide and 3.46 mg of water. What is the mass percentage of each

element? Determine the empirical and molecular formula of compound if molecular mass of compound is 88 u.

Sol. Molecular mass of sucrose = (12 × 12 + 1 × 22 + 16 × 11) = 342

342 g of sucrose contain = 6.022 × 1023 molecules

236.022 10

6.84342

= 12.04 × 1021 molecules

Number of atoms of carbon in 6.84 g of sucrose

1 molecule of sucrose contains = 12 atoms of carbon

12.04 × 1021 molecules of sucrose contains

= 12 × 12.04 × 1021 atoms of carbon

= 144.48 × 1021 atoms of carbon

Number of atoms of hydrogen in 6.84 g of sucrose

1 molecule of sucrose contains = 22 atoms of hydrogen

12.04 × 1021 molecule of sucrose contain



= 22 × 12.04 × 1021 atoms of hydrogen

= 264.88 × 1021 atoms of hydrogen

Number of atoms of oxygen in 6.84 g of sucrose

1 molecule of sucrose contains = 11 atoms of oxygen

12.04 × 1021 molecules of sucrose contain

= 11 × 12.04 × 1021 atoms of oxygen

= 132.44 × 1021 atoms of oxygen

OR

Mass of carbon present in 8.45 mg of CO2

8.45 12 mg = 2.30 mg

44

Percentage of carbon 2.30 100

54.24%4.24

Mass of hydrogen in 3.46 mg of H2O

3.46 2 mg = 0.384 mg

18

Percentage of hydrogen 0.384 100

9.05%4.24

Percentage of oxygen = 100 – 54.24 – 9.05 = 36.71%

C

H

O

54.24

9.05

36.71

Element Percentage Relative no.of moles

54.24

12= 4.52 2

4

1

Simplest wholenumber ratio

SimpleRatio

12

1

16

Atomicmass

9.05

1= 9.05

4.52

2.29= 1.97

9.05

2.29= 3.95

36.71

16= 2.29

2.29

2.29= 1

Calculation of empirical formula

Empirical formula = C2H4O

Calculation of molecular formula :

Molecular mass = 88 u





Molecular formula = empirical formula × n

= C2H4O × 2 = C

4H8O2

21. 500 cm3 of 0.300 M NaCl solution is added to 100 cm3 of 0.500 M AgNO3 solution resulting in the formation

of white precipitate of insoluble AgCl. How many moles and how many grams of AgCl are formed? Which is

the limiting reagent?

OR

Calculate number of atoms in each of the following

(i) 0.6 mole atoms of nitrogen

(ii) 0.3 mole of nitrogen gas

(iii) 3.2 g of sulphur

Sol. The balanced equation for the reaction is

AgNO3 + NaCl AgCl + NaNO

3

Number of mol of NaCl in 500 cm3 (0.05 L) of 0.300 M NaCl solution

= 0.300 × 0.5 = 0.15 mol

Number of moles AgNO3 in 100 cm3 (0.1 L) of 0.500 M AgNO

3 solution

= 0.500 × 0.1 = 0.05 mol

The reaction equation shows that 1 mol of AgNO3 reacts with 1 mol of NaCl to give 1 mol of AgCl

0.05 mol of AgNO3 will react with 0.05 mol of NaCl to give 0.05 mol of AgCl

NaCl is present in excess. AgNO3 is, therefore, the limiting reagent.

Molar mass of AgCl = 143.4 g

Mass of 0.05 mol of AgCl

= 0.05 × 143.4 g of AgCl

= 7.2 g of AgCl

OR

(i) 1 mole atoms of nitrogen = 6.022 × 1023 atoms

0.6 mole atoms of nitrogen

= 6.022 × 1023 × 0.6

= 3.6132 × 1023 atoms



(ii) 1 mole molecule of nitrogen = 6.022 × 1023 molecules

0.3 mole molecules of nitrogen

= 6.022 × 1023 × 0.3

= 1.8066 × 1023 molecules

1 molecule of nitrogen = 2 atoms

1.8066 × 1023 molecule of nitrogen

= 1.8066 × 1023 × 2 atoms

= 3.6132 × 1023 atoms

(iii) 32 g sulphur contain = 6.022 × 1023 atoms

3.2 g of sulphur contain

236.022 10 3.2

32

= 6.022 × 1022 atoms

��



Objective Type Questions

(Mole Concept)

1. A sample of ammonium phosphate (NH4)3PO

4 contains 3.18 moles of hydrogen atoms. The number of moles

of oxygen atoms in the sample is

(1) 0.265 (2) 0.795

(3) 1.06 (4) 4.00

Sol. Answer (3)

In 4 43NH PO 12 moles of 'H' are present with 4 moles of oxygen atom.

3.18 moles of 'H' are present with = 4

3.1812

= 1.06 moles of oxygen atom.

2. Which has the maximum number of molecules among the following?

(1) 8 g H2

(2) 64 g SO2

(3) 44 g CO2

(4) 48 g O3

Sol. Answer (1)

Maximum number of moles have maximum number of molecules

calculate number of moles.

8 g H2 moles =

8

2 = 4 moles 44 g of CO

2 =

44

44 = 1 mol CO

2

64 g SO2 moles =

64

44 = 1 moles 48 g of O

3 =

48

48 = 1 mol of O

3

3. The total number of electrons in 1.6 g of CH4 to that in 1.8 g of H

2O

(1) Double (2) Same

(3) Triple (4) One fourth

Sol. Answer (2)

Number of e– in 1.6 g of CH4 =

0 0

1.610 N N

16

[Total number of e in CH ]–

4

Number of e– in 1.8 g of H2O =

[Total number of e in H O]–

2

0 0

1.810 N N

18

Solutions (Set-2)



4. When x molecules are removed from 200 mg of N2O, 2.89 × 10–3 moles of N

2O are left. x will be

(1) 1020 molecules (2) 1010 molecules (3) 21 molecules (4) 1021 molecules

Sol. Answer (4)

From Equation

–(200 mg of N O)2 x molecules 2.89 × 10 moles of N O

–3

2=

=–

200 mg of N2O have molecule =

3 2320010 6.022 10

44

2.7 × 1021 molecule

2.89 × 10-3 moles of N2O have molecule = 2.89 × 10-3 × 6.022 × 1023

= 1.7 × 1021 molecules

200 mg of N2O – x molecule = 2.89 × 10–3 moles of N

2O

[2.7 × 1021 – x = 1.7 × 1021] molecule

[x = (2.7 - 1.7) × 1021] molecule

= 1021 molecule

5. Which has maximum molecules?

(1) 7 g N2O (2) 20 g H

2(3) 16 g NO

2(4) 16 g SO

2

Sol. Answer (2)

Maximum number of moles have maximum Number of molecules.

Moles of N2O =

7

44; moles of H

2 =

20

2; moles of NO

2 =

16

46

Moles of SO2 =

16

64

2H have maximum number of moles and have maximum number of molecules.

6. The number of molecules in 4.25 g of NH3 is approximately

(1) 4 × 1023

(2) 1.5 × 2023

(3) 1 × 1023

(4) 6 × 1023

Sol. Answer (2)

Moles of NH3 =

4.25

17 = 0.25 moles

Number of molecule = 0.25 × 6.022 × 1023 molecule = 1.50 × 1023 molecule.

7. The maximum number of molecules is present in

(1) 15 L of H2 gas at STP

(2) 5 L of N2 gas at STP

(3) 0.5 g of H2 gas

(4) 10 g of O2 gas



Sol. Answer (1)

Gas have maximum number of moles will have maximum number of molecules 22.4 L at STP gas = 1 mole

Moles of H2(g) =

15

22.4 = 0.67 mol; moles of N

2 =

5

22.4 = 0.223 mol

Moles of 0.5 g H2 gas =

0.5

2 = 0.25 mol; moles of O

2 10 g =

10

32 = 0.1325

2maximum number of molecules = 15 L of H (g) at STP

8. The number of atoms in 0.1 mol of a tetraatomic gas is (NA = 6.02 × 1023 mol–1)

(1) 2.4 × 1022 (2) 6.026 × 1022 (3) 2.4 × 1023 (4) 3.600 × 1023

Sol. Answer (3)

Total number of atom = 0.1 × 4 × 6.022 × 10 = 2.4 × 10 atom23 23

[4 atom are present as gas is tetra-atomic]

9. Volume occupied by one molecule of water (density = 1 g cm–3) is

(1) 5.5 × 10–23 cm3 (2) 9.0 × 10–23 cm3 (3) 6.023 × 10–23 cm3 (4) 3.0 × 10–23 cm3

Sol. Answer (4)

As water is liquid its density = 1 g/mL

i.e., 1 g of H2O have volume = 1 mL

Mass of one molecule = 23

18

6.023 10 g

23

18

6.023 10 g of H

2O have volume =

23

18

6.022 10mL = 3.0 × 10–23 mL

10. An element, X has the following isotopic composition 200X : 90% 199X : 8% 202X : 2.0%. The weighted average

atomic mass of the naturally occurring element X is closest to

(1) 201 amu (2) 202 amu (3) 199 amu (4) 200 amu

Sol. Answer (4)

Average atomic mass = percentage x atomic mass

100

= 200 90 199 8 202 2

100

= 199.96 200 amu

(Stoichiometry & Stoichiometric Calculations)

11. Two metallic oxides contain 27.6% and 30% oxygen respectively. If the formula of the first oxide is X3O

4, that

of the second will be

(1) XO (2) XO2

(3) X2O

5(4) X

2O

3

Sol. Answer (4)

This question can be solved by two methods.



wt of X8

wt of O

72.48 20.9 21

27.6

Formula of 1 st

X O3 4Oxide I Oxide II

O = 27.6%

X = 72.4%

O = 27.6%

X = 70%

X = 70%

O = 30%X O3 4

Method I Method II

Eq. mass of X =

72.4% of X = 3 mol of X Positive charge of X = =8

3

370% of X 70 2.90 mol of X

72.4 8

Atomic mass of X 21 56 g3

42

3

70Eq. mass of X 8 18.66

30

430% of O 3 4.34 atom of O

27.6 Atomic mass of X 56g

stCalculate from 1 oxide

Atomic mass 56Valency 3

Eq. mass 18.6

27.6% of O = 4 mol of O

Formula will be : X O2 3

Formula will be : X O2 3

X : O

2.9 : 4.34

2 : 3i.e.,

Similarly Atomic mass = eq. mass × Valency]

12. Haemoglobin contains 0.334% of iron by weight. The molecular weight of haemoglobin is approximately 67200.

The number of iron atoms (Atomic weight of Fe is 56) present in one molecule of haemoglobin is

(1) 4 (2) 6 (3) 3 (4) 2

Sol. Answer (1)

Weight of Fe in heamoglobin = 0.334

67200100

= 224.48 u

Mass of one Fe atom = 56 u

224.48

Total number of Fe atom = 456

13. In the reaction, 2SO2 + O

2 2SO

3

when 1 mole of SO2 and 1 mole of O

2 are made to react to completion

(1) All the oxygen will be consumed (2) 1.0 mole of SO3 will be produced

(3) 0.5 mole of SO2 is remained (4) All of these

Sol. Answer (2)

2SO + O2 2

2SO3

2 moles of SO2 reacts with 1 mole of O

2 as 1 mol of SO

2 is present

SO2

will be limiting reagent will formed 31 mol of SO



14. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in

this reaction will be

(1) 1 mol (2) 2 mol (3) 3 mol (4) 4 mol

Sol. Answer (4)

2 2 22H O 2H O

4 g 32 g 2 moles

As 10 g of H2 is to react with 64 g of O

2

O2 will complete consumed and will act as LR.

Calculation will be made on the basis of weight of O2

2 2

2 2

32 g of O gives H O = 2 moles

64 g of O gives H O = 4 moles

15. Consider the following reaction sequence:

S8(s) + 8O

2(g) 8SO

2(g)

2SO2(g) + O

2(g) 2SO

3(g)

How many grams of SO3 are produced from 1 mole S

8?

(1) 1280 g (2) 960 g (3) 640 g (4) 320 g

Sol. Answer (3)

1 mole S8 8SO

3 8 × 80 g = 640 g

16. What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1 L of propane

gas (C3H

8) measured under the same conditions?

(1) 10 L (2) 7 L (3) 6 L (4) 5 L

Sol. Answer (4)

C H + 5O3 8 2

3CO + 4H O2 2

For 1 mol propane 5 mol O2 gas is needed.

22.4 L propane = 5 × 22.4 L of O2 gas needed

21 L propane = 5 L of O gas is required

(Reactions in Solutions)

17. 4 g of hydrogen reacts with 20 g of oxygen to form water. The mass of water formed is

(1) 24 g (2) 36 g (3) 22.5 g (4) 40 g

Sol. Answer (3)

2 2 22H O 2H O

4 g 32 g 36 g

When 4 g of H2 reacts with 32 g of O

2 gives 36 g of H

2O.

Now present oxygen is 20 g

O2 will be the limiting reagent and H

2O will be calculated from O

2

32 g of O2 given = 36 g of H

2O

20 g of O2 given =

3620

32 = 22.5 g H

2O

18. Calculate the molality of solution containing 3 g glucose dissolved in 30 g of water. (molar mass of glucose

= 180)

(1) 0.50 m (2) 0.56 m (3) 0.091 m (4) 0.05 m



Sol. Answer (2)

Molality = B

A

nmoles of solute

wt of solvent (kg) w (kg) w

B = 3 g, w

A = 30 g

Molality (m) =

3

1801000

30 =

11000

1800 = 0.56 m.

19. How many grams of NaOH should be added to water to prepare 250 ml solution of 2 M NaOH?

(1) 9.6 × 103 (2) 2.4 × 103 (3) 20 (4) 24

Sol. Answer (3)

Moles of NaOH = M V(mL) 2 250

1000 1000

= 0.5 moles of NaOH

Moles = given mass

mol.mass 0.5 mole =

x

40 given mass = 40 × 0.5 = 20 g

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