u· /J

EXERCISES

1. How many moles of atoms are there in o~e atom? 2. Calculate the number of moles in

(a) 315 g of Mg(HCO3)2 (Mg = 24, C = 12) (b) 3·6 mg of NH3 (N = 14) .

3. An unknown gas occupies 322 mL at 29°C and 616 mm. How many molecules are there in this gas ?

4. What is the weight of one millimole of S 02 ?

. S. Calculate the weight of 3 · 491 x 1019 molecules of Cl2.

6. If one atom of an element X weighs 2 · 107 x 10-22 g , what is the atomic weight of x ? 7. How many atoms of hydrogen are there in 2 · 57 X 10-6 g of hydrogen ? 8. The total population of the world is now believed to be about 4. 2 x 109 pe 1

How many moles of people are this ? If you had one sulphur atom for each · e~p e. what would be the weight of the sulphur sample 1 P son,

9. How many moles, molecules and atoms of pho,phorus are contained in 92. 93

f phosphorus, if the formula of its molecule is , 4 ? g o

10. o. 44 g of a colourless oxide of nitrog~n occupies 224 mL at L520 mm of H d ., Identify the compound. How. much m grammes docs one molecule of ti! an 273 C . .

igh ? . compound we .

11 From 200 mg of CO2, 1021 molecules are removed. How many mole f C

· s O 02 are left ?

l

THE MOLE CONCEPT 33

12. Calculate the volume occupied by on _. • . · . water be 1 · 0 g/mL. e molecule of water assummg the density of

13, The density of mercury is 13 · 6 g/mL C· 1 1 . . assuming that each atom of mer , · . a cu ate the diameter of an atom of mercury

cury is occupyin , b f d J th 1 h diameter of the mercury atom. (Hg == 200) · g a cu e o e ge- eng equa to t e 14. How many g-atoms of hydrogen are present in (i) 8·5 f NH (" ) 0 02 [11.lT/f CuS04. 5H20? go 3, 11 • mo e o

15. How many moles of oxygen gas that b b • tion of 90 g of water ? can e O tamed by the electrolytic decomposi-

16. How much calcium (in g) is present in Ca(N03) that co t . 1 4 . [B?.l T Mesra] . 2 n ams · g mtrogen . 17. Assume that the n~cleus of flu~rine atom is a sphere with a radius of 5 x 10-13 cm.

Calculate the density of matter m the fluorine nucleus. (F = 19) /RoorkeeJ 18. Calculate how many methane molecules and how man h d d b t are there in 25 · 0 g of methane ? Y Y rogen an car on a oms 19. Calculate the number of ato~ in l·0 mg of helium gas and 1 f 1.0 [JSf.Mtl.J gas at S.T.P. (He = 4) _. vo ume o g o 1e

20. Calculate the total number-of electrons in ·1 • 6 ~ of methane.

.. • _. . . . . · . . /llT, TS. Rajendra] 21. A signature wn_tten ~ Carbon pencil weighs 1mg. What is the number of Carbon

atoms present in the sigbature ? , . (Kamataka Engg. )

22. You have 0 · 4 ~ole of H 2S in bott_le. Calculate · (i) the number of grammes of H ].5, (ii) number of gr~es-of hydr'ogen and sqlphtir, (iii) number of molecules of H2S and (iv) number of atoms 9f hydn;>g~n and sulphur present (H = l · 00; S = 32 • 0).

- \ [B.JTJ 23. A certain sample pf nitrogen gas cpnsists of 9 · 26 x 1022 nitrogen atoms.

(a) How many moles of N atoms arepre_senffu:this saritple? . (b) if the gas is entirely in molecular form, how= many. molecules are present · in this

sample? ( c) What is the mass of the sample in gramme ? f B.l T J

24. DNA is thought to be a chemical compound respo-nsible for the process of heredity. Calculate the average volume occupied by one molecule of DNA if its density is 1.1 g/mL and its molecular weight is estimated to be 6 x 103 gi'mole. · [B.JTJ

25, Calculate the number of ions in 1.4 mg of N3-.

26· Find the charge of 27 g of Af + ions in coulomls. 27. How many years would it take to spend Avogadro number of rupee at the rate of 1 million

rupees per second ? 28. A is a binary compound of a wuvalent metal 1 · 422 g of A reacts completely with 0 · 321 g of

sulphur in an evacuated and sealed tube to give 1 · 743 g of a whitt: crystalline solid, B, that forms . a hydrated double salt, CwithAl2(SO4)3. Identify A, B andC, (S = 32· 1) fllTJ 29. An alloy contains A, B and C atoms of metals in the ratio of 2 : 4 : 3, respectively.

What is the mass of the sample of this alloy containing a total of 4· 5 x 102

atoms ? (At. wt. of A, B, C is 50, 75 and 100, respectively) . . . . .. ·

30. Chlorophyll the green colouring matter of plants contains 2.68% of magne.sium by weight. Calculate the number of magnesium atoms in 2.00 g of chlorophyll (at mass of Mg= 24).

1j

I

34 ~UMERICAL EXAMPLES IN CHEMISTR y

· i d MN2 and 'My"lz. 31. Suppose two elements Mand N combine to form two compoun 5 . , . . h 18 Calculate atomic 0 · 05 mo.e of MN2 weighs 5 g while O • 1 mole of Mf'/2 weig 5 g.

weights of M and N.

OBJECTIVE QUESTIONS

(A) Only one answer is correct : Wh' h of them will have the 32. 5 ·0 g each of the following gases at 87°C and 750 mm pressure are taken. ic

least volume ? · (a)HF (b) HCl (c)HBr (d)Hf.

33. One mole of fluorine gas :

(a) weighs 19 ·0 g

( c) contains 1 · 20 x 102~ F atoms

(b)contains6 ·02 ~ 1023

Fatoms

( d) weighs 6 · 02 X 1023

g. 34. 10 · 0 g of CaCO3 contains :

(a) l ·0 mole of calcium (b) 10 moles of~aCO3

(c)6 ·<!_2 x 1022 atoms of Ca (d)6.02 x 2023 atomofCa. 35. Which of the following has the largest number of atoms ? ·

(a) 0·5 g-atom of copper (b) 0•635 g of copper

( c) 1 x 1a23 atoms of copper. ( d) 63.5 mg of copber 36. A gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 b}'. weight. The ratio of number of

molecules of oxygen and nitrogen is : (a)1;_4 (b)2 : 7

37. At O"Cud 0 · 5 atm, 4 ·48 litresNHJ gas: (a) contains 0·20 moleNH3

(c) 7: 32 (d) 3: 16.

{b) weighs 1 · 70 g

( c) contains 6 · 02 x 1a23 molecules (d) contains 0·40 moleNH3. 38. The weight of millimole of water in gramme is :

(a)1· 8_g . {b)0 ·18g {c)1·8Xl0-2g { d) l · 8 X 102 g. 39. Avogadro number is the :

(a) weight of any substance in -gram (b) weight of one litre of a gas at S.T.P.

( c) weight in gram x 6 · 02 x 1023 ( d) number of molecules in a mole.

40. Which of the following contains the same number of atoms as 6 g of magnesium ? (a) 12 g of cabon (b) 4 g of oxygen gas (c) 6 g of ozone gas {d) 27 g of aluminium.

41. The molecular mass of each of N2 and CO is 28. If 0 · 5 L of N2 at 27°C and 700 mm pressure contains n moleculei., the number of molecules in 1 · 0 L of CO under identical conditions will be : (a)n/2.. (b)n (c)2JJ {d)none.

42. When the same amount of zinc is treat~d ~eparately with excess of sulphuric acid and excess of NaOH the ratio of the moles of hydrogen evolved 1s : , (a) 1 : 1 (b) 1 : 2 ( c) 2 : 1 ( d) 2 : 3.

43 A -gas· eous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight There~o h . • I . . . ,, re, t e ratio of their number of molecu es is. (a)l :4 (b)l : 8 (c)7 :32 (d)3 : 16 Th se of 12c; scale has superseded the older scale of atomic mass based on 160 isot . . .

44. ad:a~tage of the former b1tng ope, one unportant ic masses on C scale become whole number .

(a) ~~e-~tom abundant in the earth crust than 120 . -(b) C 1s more . . d'ff ence between the physical and chemical atomic masses got narrowed down . ·c· (c) the 1 er . s1gn1 1cant1y 12 . • t d midway between metals and non-metals In the periodic table (d) C 1s s1tua e ·

THE MOLE CONCEPT / 35

Which of the following weight the least 'f 45· . ) 2gatomsofN(at.wt. ofN = 14) (b)) 23

(la) 1 mole of S (at. wt. of S = 32) (d) 7 X ~O atoms of carbon (at. wt. of C = 12) c g silver (at wt of Ag -108)

19 7 kg of gold was recovered from a person. How man · · -46, · (b) 6 02 x 1023' Y atoms of gold were recovered?

(a)lOO · (c)6.02xto2~ (d) 602 25

47. 4.o gram pf ca~stic soda c~natians . x 10

, (a) 6.02 x tt atoms of It (b) 4 g atoms of Na '\ (c) 6.02 X 10 atoms of Na ( d) 4 moles of Na OH

48. ) Which of the following weight the least? 1

{a) 24 gram of magnesium (b) 0.9 moles of nitric oxide (c) t2.4 litre of N2 . ( d) 6.02 x 1024moieculcs of oxygen

49. Thdargest number of molecules is in

(a) 54 g nitrogen peroxide (b) 28 g of carbon monoxide (c) 36 g of water ( d) 36 g of ethyl alcohol

so. _The molecular_weight of 02 and S02 are 32 and 64 respectively. If one litre of 02 at 1s0 c and 750 mm pressure contains N molecules! the number of molecules in two litre of so2 under the same conditions of temperature and pressure wtll be · · . ~ · . . . -

N OOi MN WW ~w f SL One mole of CO2 contaim

{a) 6.02 x 1023

atoms of C (b) 6.02 x 1023 atoms of 0 ' 23 -

(c) 18.1 X10 molecules of CO2 ( d) 3 gram atoms CO2

51.. H 0.5 mole...Df BaClz is mixed with o.2 rnol of Na'j/'04, the maximum number of mol of Ba3(PO4)2, that can be f ormccps .(a)0.7 ' · (b)05 (c)0.30 (d)0.10

51 2ig of Al will react complet~ly with how many grams of oxygen? (a) 8g _ {b) 16g (c) 32g (d) 24g

- St. How many molecules of acetylene would be formed when 10.0 g of calcium carbide is treated with water? (A~rdro's constant = 6.02 x 10~/mole; Ca = 40, C = 12, 0 ; 16 and H = 1)

- ·(a)9A_x_ 1022 (b) 94 X 1020 . (c) 9.4 X 1023 - (d) 18.8 X 1022

·ss. The relative atomic mass of an element is 58. Which of the following samples will contain the least number of atoms ? ·

(a)58molcs {b)580a:mu (c)5.8 x 10-10g . (d)0.0058g (PaC.E.T.)

5'. 0.24 g of a volatile compound ui:on vapourisation gives 45 mL vapour at NTP. What will be the vapour density of the substance? (Density of H2 = 0.089) (a) 95.39 (b) 5.993 (c) 95.93 (d) 59.73 (CBSE.) .

57. How many ':lfen atoms are present in 0.5 mole of su~hur dioxide? (a) 6.02 X 10 (b) 3.01 X 1023 (c) 9.03 X 10 (d) 12.04 x 1023 (Model Paper PUCET.)

51. Volume occupied by 4.4 g CO2-:it S.T.P. is (a)2.24Jitres (b)2-2.4litres (c)224litrcs (d)noneofthese (AFMC.)

59. Haemoglobin contains 0.33% iron by weight. This molecular' weight of haemogobin ii ,..approximately 67200. The number of iroMltoms (at. wt. of iron is 56) present in one molecule of haemoglobin are (a)l (bJ6 (c)4 {d)2 (CBSE.)

60· The number: of gram molecules of oxygen in 6.02 x 1024CO moleules is .(a) 5 gram molecules {b) 8 gram-molecules ( c) 2 gram molecules _ ( d) 0.5. gram n.iolecules ---tPb. C.E. T •. )

61.. The numb.er of molecules in 67.2 L of gas at 0°C and 1 atm pressui'~ •_is · 23

(a) 6.02 X 1023 {b) 12.04 >< tCf3 (c) 18.06. X 1023 (d) 24.08. X 10 (P.U.C.E. T. ~-qdel Pa~r)

62. Number of g of'oxygen i-n. 32.2g Naz SO4J0H2O is : (a) 20.8 (b) 22.4 (c) 2,24 (d) 2.08 .

~~-'>~;';>,~1.:i=..-n-,,:.s;-:e,-•, n •• . ~ --·-· -

fr

36 NUM ERI CAL E XAMPLES IN CHEMISTRY

0 0018 mL) at room tem perature is : 63. The number of water molecules present in a drop of water (volume ·

19 (a) 6.023 X 1019 (b) 1.084 X 10

(c) 4 84 X 1017 (d) 6.023 X 1023

. -3 · . . . . .J d h t of water vapour 1s 0.0006 g r.m ,

64. At 100°C and 1 atm, if the density of hqu1d water 1s 1.0 g cm an t a rature is: then the volume occupied by water molecule in 1 litre of steam at that tempe .3

.3 .3 ( ) 0 6 cm·3 (d) 0.06 cm (a)6cm (b)60cm c · L f05MCa(OH)2willb'.!:

65. The mass of CaC03 produced when carbon dioxie is bublled through 5oo m O

· (AFMC)

(a)lOg (b)20g (c)50g (d)2Sg . . · -2 / shose radius and length are 7A and

66, Specific volume of a cylinderical virus particle 1s 6.02 x 10 cc gm

10A repectively. If NA = 6.02 x 1023, find molecular weight of virus. 3

(a) 15.4 kg/mol (b) 1.54 x 104 kg/mol (c) 3.08 X 104 kg/mol (d) 3.08 X lO_ kg/mol 1 les produced 1s equal to:

When 0.1 g of hydrogen is brunt in oxygen, the number of water mo ecu 22

(a) 6.02 X 1023 (b) 3.01 X 10~ (c) 3.01 X 1022 (d) 6.02 X 10 (Pb.C.E.T.) 67.

How many moles of electrons weight one kilogram?

23 1 · 31 . ) 6.023 1054 (a) 6.02 x 10 (b) 9.108 x 10 (c 9.108 X

68. 1 X 108

(d) 9.108 X 6.023 (IIT.)

1-1) . Number of atom in 558.5 g Fe (at. wt. of Fe = 55.85 g mo , 1s

22 (a) twice that in 60 g carbon (b) 6.023 X 10

23 ( c) half that in 8 g He ( d) 558.5 X 6.023 X 10

3 • • • 0.56 g of a gas occupies 280 cm at SfP, then its molecular mass IS ...

(a) 4.8 (b) 44.8 · (c) 2 -

69.

70. (d )22.4 (Orissa J.S.E.)

The maximum number of molecules will be present in

(a) 16 g of N02 gas (b) 16 g of 02 g~s (c) 7 go( N2 gas 71.

(d)2gofH2gas (CBSE.)

72. Which has maximum number of atoms ?

(a) 24 g C(l2) (b) 56·g of_Fe(56) (c) 27.g of (Al(27) _ (d) 108 g of Ag(108)

7 3. The litres of CO2 represented by 4.4 gm of ·co2 at SfP are :

-(IJT.'·

.(AFMC.) (a) 2.4 litre (b) 2.24 titre (c) 44 litre (d) 22.4 litre

74. Maximum number of molecules are present in: ·

75.

76.

77.

78.

79.

80.

(a) 15 Lof Hz gas at SfP (b) 5 L of N2 gas at SfP

w~~~~gas ~10p~~gas (CBSEMed. )

An alkoloid contains 17.28% of nitrogen and its molecular mass is 162. The number of nitrogen atoms

present in one molecule of alkaloid as

(a) five (b) four (c) three (d) two (Kerala PMT.)

For the formation of 3.65 p of hydrogen chloride gas, what volume of hydrogen gas and chlorine gas are

rqeuired at SfP conditions? _ ·

(a) 1.12 litre, 1.12 litre (b) ·l.12 litre, 2.24 litre

(c) 3.65 litre, 1.83 litre (d) 3.55 litre, 1.83 litre (Kerala PMT.)

If we consider that ¼ in place of 1/12 mass of cargbon atom is taken to be relative atomic mass unit, the

mass of one molecule of a substance will

(a) increase two fold (b) decrease twice

(c) be a function of molecular mass of the substance (d) remains unchanged

If 30 mL of hydrogen and 20 mL of oxygen react to form water what is left after the d f h . en o t e reaction?

(a)lOmlofH2 (b)SmlofH2 (c)10mlofO2 (d)5mlofOz .

f rb d d ( · · 1 . (AFMC.) The mass o ca on ano e consume g1vmg on y carbon dioxide) in the p od . · aluminium metal from bauxite by Hall's process is: r uchon of 270 kg. of

(a) 90 kg (b) 540 kg (c) 190 kg (d) 270 kg

Mixture X = 0.02 mo! of !Co(Nlh)sS04]8r and 0.02 mol of [Co(NH3) 5 Br] SO (C~SE Med.) • 4 was prepared In 2 litre of

solution 1 litre of mixture X + Exces of AgN03 ➔ y

THE MOLE ~ONCEPT 37

. of mixture X + Execs of BaC/i ➔ z 1wc

bcr of moles of Y and Z a re : Num ( )

O.Ol , 0.01 . (b) 0.02, 0.01 ( c) O.O l a ' O.Q2 ( d) 0.02, 0.02

(8) More than one choi'c . e 1s correct

Which of the following con st i I u te 0.1 g mole ? 1, . 23

(-) 6 022 x 10 molecules of benzene (b) 0 14 a · , , · g of N2 gas

( ) 2 24 htre of CO2 at S.T.P. (d) o 40 c · · g of He gas R k

" II · h I ( oor ee)

2. Which of the ,o owing ave equa number of molecules?

(a) 0.5 mole of CO2 gas (b) 11.2 L of CO2 gas at S.T.P.

(c) 22 g of COz gas _ ( d) 6.022 X 1023 molecules of CO2

3 Ch~C the correct statements :

' 22 . (a) 1 mole means 6.023 x 10 particles

(b) Molar mass is mass of 1 mole of a substance

(c) Molar mass is mass of 1 molecule of substance

(d) Molar mass is molecular mass of a substance expressed in g.

4. Which of the following have same number of atoms ?

(a) 0.1 mole 02 (b) 0.2 g atom of He

(c) 0.1 mole of P4 (d) 0.2 mole ofSs

5. The amount of oxygen that can be obtained from 90 kg of water is .

(a) 90 kg (b) 80 kg ( c) 2.5 g molecule ( d) 2.5 kg molecule

(C) Objective Questions b_ased on Compreh~nsion

[A] The term mole first used by Ostwald in 1896 refirs for ·the ratio·· of mass of a ~ubstance in g and its -

molecular we~ght: 1 mole_ of a ~aseous _compound occupied 22A L-aFS·.T.P. and-contains 6.022 x 1023 . .

molecules of gas : · ~- --- · _ _ _ · -_ _ , - -.

[B]

l. Weight of 1 atom of hvdrogen is

(a) 1.66 X 10!4 u -· _ _(bf 3.~2 -X 10-,24 g ,_ ,

( c) 1.66 X 10-24 g ( d)3.32 ·x -10.-24 u _; -

2. Avogadro's number of Rupees can be spent in years ........ '. .. : ..... if 10 lacs rupees per second are ~pent

(a) 1.91 x 1010 years .(b) 2.91 x tol~ year

(c) 3.91 x 1010year (d) 4.91 x 1010 y~a~ ,

3. The amount of sulphur required to produce 100 mole-of H'JS04 is

. (a) 3.2 X 103 g (b) 32.65 g

(c)32g (d)3.2g

4. The vapour density of a mixture containing N02 and N204 is 38.3 at 27°C.. The moles of N02 in 100 mole

mixture is

(a)33.48 (b)53.52 (c)38.3

S. The volume of air ·qeeded to bum 12 g carbon comple(ely at S.T.P. is

(a)22.4L - (b)112L (c)44.8L

6. The maximum number of atoms present are in

ta)4gHe (b)4g02 (c)4g03

Paragraph ,/ .· ·

(d) 76.6 ·

(d) 50 L

[IIT.]

Chemical reactions involve interaction of atoms and molecules. A large number of atoms / molecules

(approxim~tely 6.023 x 1023) are present in a few grams of any chemical compound va1ying with their

atomic / molecular masses. To handle such large numbers conveniently, the mole concept was intro­

duced,. The concept has applications in diverse areas such as analytical chemistry, biochemist1y,

electrochemist.ry ai1d radiochemistry. The following example illustrates a typical case involving chemi­

cal/(;(ectrochemical reaction which requires a clear understanding of the mole concept.

38 NUMERICAL EXAMPLES IN CHEMISTRY

A 4.0 molar solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes. (atomic mass : Na = 23, Hg = 200; 1 Faraday = 9~500 Coulombs)

1. The total number of moles of chlnrine gas evolved Is (a)0.S (b)l.0 (c)2.0 . (d)3.0

2. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from th is solution is (a) 200 (b) 225 (c) 400 (d) 446

3. The total charge (Coulombs) required Cor complete electrolysis is (a) 24125 (b) 48250 (c) 96500 (d) 193000

(D) True and False State whether the following statements are aue or false ?

1. Avogadro constant is designated by NA and has have 6.022 x 1023 mor 1

2. One mole of chlorine contains two gram atoms. 3. One mole of electrons contains 6.022 x 1022 electrons 4. One gram molecular weight of every gas occupies 22400 mL at S.T.P. S. The molecular weight of a compound is just the weight of a molecule.

(E) Fill in the Blanks l. The number of molecules contained _in 8 g of oxygen .... .... 2. The molecular weight of chlorine is 71. One molecule of chlorine weights ......... . 3. The number of g atoms contained in 5 g of Ca is ............ .. 4. The number of molecules present in 7 g of CO is .. .. ...... .. .. S. The volume occupied by 2 g of H2 and 32 g of 02 at STP is .............. cm3

. (F) Match the Column I with II

M atc.h the entries of colwnn A with appropriate entries of column B. I II

L 22.4 Lat STP. (a) 108 g _ 2.:1 g atom of rhombic sulphur (b) 44 g CO2 3. 1 g atom ofAg (c) 0.1 mole

1 4. 1.8gHz() (d) 8gmo\ccule

(G) Assertion-Reason 1ype Questions

The Que,tions given below consi~t of an Assertion in Column l and the Reason in Column 2. Use the f ollOM key to choose the appropnate answer. (a) If both a55ertion and reason arc correct and reason is the correct explanation of the assertion. (b) If both a'"rtion and rea50n are correct, but reason is Nor nm CORRECT explanation of the assertion (c) If assertion is correct but reason is incorrect. (d) If both assertion and reason are INCORRECT.

1. A : 1 g atom of ,ulphur c~nt~ins molecules equal to Avogadro constant R : Atomicity of 1ulphur 1, eight

w ~ . w . ~ Z. A . The pumber of oxygen atoms 1n 1 g of Oz, 1 g of 03 and 1 g of atomic oxygen is same ' 1 R : Each of the species represent 16 g atom·of oxygen

(a) (b) (c)

coulombs)( d)

THE MOLE CONCEPT

. 'fhc formula of iron (III) ferrocyanide is Fe~ [Fe(CM ] .3 A. . . ,. 16 3

· R: 860 g of iron (III) ferrocyan1de contains 7 iron atoms

(a) (b)

(c) (d) . 1 Avogram is equl'II to 10 amu

4. A , I . I . . R: 1 amu is cqau to x g w 1ere x 1s rcc1procal of Avogadro number

W M (c) (d)

s. A: 1 g atom of iron represents the number of iron atoms present is 1 g of it.

R: 1 g atom of element weights equal to GAM of element

(a) (b)

(c) ( d)

.HINTS FOR HARDER PROBLEMS

13. The length of the side of cube = diameter of mercury atoiu = x cm

:. volume of the cube = volume of the atom of Hg = x3

f mass oLone atom .. volume of one atom o Hg = d · . · em~

x3 = 200 ·x _1_· 6 · 02 X 1023 13 ' 6

Ta~g logrithm on both sides and solving, x = 2 · 9 x 10-8 cm.

39

20. Learn that one-molecule of methane consists of one carbon atom and four hydrogen

atoms. Hence the number of electrons in one molecule of methane.

· = _6 + 4 = 10 electrons

The number of CH4 molecule in 1; 6 g methane

1·6 = no. of mole of methane X NA =

16 X NA

. . . 1·6 · No. of electrons m 1 · 6 g methane =

16 x NA x 10.

24. Mass of one molecule= (6 x 103)/(6 x 1023) = 10·20 g

l f . 1 mass of a molecule 10- 20 . L 9 1 x 10- 21 L vo .o amo·~cue = . . = - - m = · m.

: density 1 · 1

28. Univalent inetlil is an alkali metal. (A) is a compound of an alkali metal and another

element. A + s B

1 ·422g 0·321 g 1 ·743 g

0·321 No. of mole of S =

32_ 1 mole = 0·01 mole

(B) forms a hydrated double salt withA/2(SO4)3. (B) is, therefore, most likely K2so4.

Note the formula of potash .alum. Mol. wt. of K~P4 = 39· 1 X 2 + 32· 1 + 4 x 16

= 174 · 3. Mole of K~ 04 fanned = 1 · 743/174 · 3 = 0 · 01 mole. From reactior. it appe~s

that (A) is ~ compound of potass ium and oxygen. One mole of (B) ccntaim four

40 NUMERICAL EXAMPLES IN CHEMISTRY

moles of_ oxygen atoms which must have;bceen provided by (A) ~one. No compouocl of potassmm and oxygen ha\~ng formula (femove S from K2SO4) eX1Sts. Hence the com-pom1d is most likely KO2, and its two moles must have reacted with one mole of sulphur atom. Mol wt. of KO2 = 39 · 1 + 32 = 71 · l g/molc. Wt. of (A) corresponds to 0.02 mole of (A). Hence the reaction is 2A + S - B 0·02 mole 0·01 mole 0·01 mole (A) =KO2, (B) =K2SO4 and (C) =K2S04·Al2(S04)3·24H20

29. Calculate atoms of A, B and C and then the mole of each metal. Multiply mole of each metal by its atomic weight, add together. Do the rest. 31. t1oL wt. of A!N2 = 5/0 · 05 = 100 gi'mole

Mol. wt. of Al~2 = 18/0·01 = 180 g/mole Suppose at. wts: of Mand N arex indy, respectively. · For AfN2, x + 2y = 100, and for M~2, _ _ 3x + 2y ~ 180 Solve for x andy

Hints For 'fypical M-µl!ip!e Ghoic~ Questions 52. 3BaCl2 + 2Na:,l'04 ➔ Ba3(P04)2 + _6Na'c1_-

3mol

0.3 2mol

0.2mol ·

1 mol - . -- : · 0.1 m1r ~ :- ' ' :\:_:~ -~ -

:.C,-c,7-::. '=:...

BaCli (0.5 mol) is in excess_. H~nce·fa_a_:,Po4· gets fully consumed which is the limit-ing reactant concept. - - -_ · - -_ - -- -,_.: - ~ - - - - -Therefore, the amount of B-~3(PQ_i)i -for:~~~·:is:cal~ulated from it.

54. CaC2 + 2H2O-+ Ca(OH)2 +C2H2 .· ,- __ , ·: ~-40 + 12 x 2

64 g 26 g-26 l0gCaC2 = 64

x l0gC2H2

Jg 10 No of moles of C2H2 = 64 x :;kt'

. 10 No of molecules of C2H2 =

64 x 6.022 x 1023

= 60.22 X 1023 64 .

= 9.4 X 1022

56: 45 mL at S.T.P. = 0:24 g

P g,24

22400 mL at S.T. . = 4S X 22400"g

Mol. wt. = 119.47

Mol. wt. - 11~47 = 59_73 · V.D. == 2

THE MOLE CONCEPT

r:e present in 67200 amu = Ol.C3~ X 67200 == 232 - 222 59, ri J amu - - == 4 atoms.

67.2 .,. 56

61. No. of moles of the gas = 22.4 = 3.0

:. No. of molecules = 3 x 6.022 x 1023

= 18.06 X 1023

62, Mol. mass of Na2S04. lOR20 == 2 x 23 + 32 + 4 x 16 + 10 x 18 = 322 :·. 322 g of Na2S04 . 10 H20 contains oxygen == 4 x 16 + 10 x 16 = 224 g

322 g of Na2S04. 10 H20 contain of oxygen = ~~~ x 32.2 = 22.4 g

63• Mass of one drop of water = vol. X density = 0.0018 g

Number of water molecules in 18 g water (1 mole) = 6.022 x 1023

Number of water molecules in 0.0018-g water

6.022 X 1023 19 - 18 X 0.0018 = 6.022 X 10

64. Weight of water vapour in steam = 1000 x 0.0006 = 0.6 g .

41

Sine the liquid water has density 1.0 g cm - 3 , the volume occupied by water molecules is ·

0.6cm3

65. Ca(OH2) +CO2 ➔ CaC03 + H20; 0.25 ~ol. CaC03 = b.25 x 100 = 25 g

0.25 mol _Q.25 mol . 500

No. of moles of_C[!(OH)2_ = 1000

X 0.5 M = 0.25

66. Specific volume (volume/g) o( cylindrical virus = 6.02 x 10-2 mL/g

Radius of virus (r) = 7A ~ i x\ o-8 c~ .. ·

Volume of virus = nr2 I (/ = length of Vitus) ·

= 22

X (7 X 10-8)2 X 10 X 10-8 = 154 X 10~23 mL. 7

volume 154 x 10-23

Weight of one vii us particle = 1 = 6 02 10

_2 \ sp. vo ume . x

Molecular weight of virus = Weight of 6.02 ; 1023 particles 23 \

= 154 X 10- X 6.02 X 1023 = 15400 g/mol = 15.4 kg/mol 6.02 X 10-2

68· Mass of an electron= 9.108 x 10-31 kg

. ·. Mass ·of 1 mole of electron 10-8

= 6.02 X ~ . X 9.108 X r° kg

= 6.02 X 9.108 X 10-8 kg

:. No of moles in 1 kg 0 of electrons

,.-1

42

I

NUMERICAL EXAMPLES IN CHEMISTRY

1 10~ = 6.02 X 9.108 X 10- B .= 6.02 X 9.l08

· 0,56 X 22400 - 44 8 g 70. 280 mL gas weighs =-0.56 g; 22400 mL gas weighs = 280 - ·

. 7 . . 71 10 g NO = 16 = 0 35 l · 15 O = 16 = 0 5 mol · 7 g N2 = - = 0,25 mol , 2 g of • 2 46 . mo ' g 2 32 . , 28 H2 = 1 mol.

72. (a) 24 g C = ~~ = 2 mole carbon. All others are 1 mol each

73. (b) 4.4 g CO2= :: = 0.1 x 22.4 = 2.24 litre

15 74. (a) 15-LH2 at S.T.P. = 22

.4 mole= 0.67 mole

' 5 (b) 5LN2 atS.T.P. = 22

.4 = 0.22mole

(c) 0.5 g of H2 = 0; = 0.25 mole

10 ·. ( d) 10 g 02 = 32

= 0.32 mole.

7 5. ( d) 100 g of alkaloid contain nitrogen = 17 .28 _g 162 g of alkaloid contian introgen =

17·~

0~

162 = 27.99 = 28 g

or one mole of alkaloid contain 28 g nitrogen or ~ = 2 g atoms of nitrogen :. Number of nitrogen atoms in molecule of alkaloid ..... 2

76. (a) H2 + C/2 -- 21-ICI Hydrogen chloride formed= 3.65 g = ~; = 0.1 mol 0 .5 mole 0.5 mole 1mole -0.05 mole 0,.05 m~le 0.1 mole Volume of H2 = 0.05 X 22.4 litre, volume of C/2 = 1.12 litre

78. (d) 2H2 + 02 -- 2H20 2mL lmL

30 mL 15 mL 'F~r 30 mL of H2, the volume ofoxygen consumed = 1.5 mL .·. Volume of oxygen lyft = 20-15 = 5 mL 108 g of Al is produced when carbon = 36 g

270 g of Al is produced when carbon= 1~~

4 x 270 =90 g

THE MOLE CONCEPT

0 + 3 C ➔ 4 Al + 3 CO2 .19• 2Al2 3 3x12 4 >< 27

4 >< 27 kgAl == 3 x 12 k~ C

3 xg 10

:. 270 kgAl .== 4' X2 x ~ kg C == 90 kg of C.

ANSWE RS

1.· 1 · 66 >< 10- 24 mole 2. (a) 2· 15 mole (b) 2· 1 X 10-4 mole

3. 6 · 34 x 1021 molecules

4. 6 · 4 X 10-2 g 5. 4 · 11 7 mg 6. 126 · 8 g/mole

7, 1. 53 x 1018 atoms . 8. 6 · 9 X 10-15 mole, 2 • 2 x 10-13 g . 23

9. o. 75 mole, :t · 51 X 10 molecules, 1 · 804 x 1024 atoms

10. N20 ; . .. 11. 2 · 88 X 10-3 mole 12. 3 · 0 x 10-23 mL

7 · :-1 X 10-23 g . ')

l3. 2·9 x 10-.1 cm

16. 2·0 g

14. (i) 1 · 5 g-atom (ii) 0 · 2 g atom

17. 6·025 X 101'.i g/mL

15. 2· 5 mole

18. 9 · 40 >< 1023

molecules CH4

43

19. 1 · 5 x 1020 atoms, 5 · 6 20.' 6 · 02 x 1023 electrons

litres 21. 5 · 02 x 1019 atoms of C

22. (i) 13 · 6 g , (ii) H = 0 · R g S = 12 · 8 g

(iii) 2 · 4 x 1023 molecules ·

(iv) H = 4.81 x 1023 atoms S = 2 · 408 x 1023 atoms

24. 9 · 1 X 10-21 mL 25. 6 • 022 X 1019

23. ( a) 0. · 153b mcil~ (b) 4 · 63 X 1022

molecules (c) 2· 15 g

26. 2 · 89 x 1a5 coulombs

27. 1 · 9 x 1010 years 28. A =K02 , B =K2S04

C=K2S04 ·Al2(S04)3. 24H20

29. O· 581 kg 30. 1·3 x 1021 atom 31. M = 40 N = 30

OBJECTIVE QUESTIONS

(A) Only one answer is correct

32.d 33.c 34.c 35. a 36.c 31.b

38.c 39.d 40.b 41. C 42.a 43.c

44.c 45.b 46.d 47.c 48.a 49.c

50 .. c 51. a 52.c 53.d 54. a 55.b

56.d 51.a 58.a 59.c 60.a 61. C

62.b 63.a 64 .c 65.d 66.a 61.c

68. d 69.a 10.b 71.d 72. a 13. b

74.a - 75.d 16.a 77.b 78.d 79. a ~o a.

44 NUMERICAL EXAMPLES IN CHEMISTR y

(B) More than one choice is correct

1. c, d 2. a, b, c 3. b, d 4. a, b 5. b, d

(C) Objective Questions based on Comprehension

[A] 1. c 2. a [B] 1. b 2. d

3. a

3. d 4. a 5. b 6. a.

(D) True and False

1. True 2. True · 3. False 4. True 5. True

(E) Fill in the-Blanks

..

1. 1.5 X 1023 2 118. 10-22 .· • . x_~ ' -_ ~ g 3~ 0.125 4. 1.5 X 1023

(F) Matc_h the Column I with II

1. ➔ b 2. ➔ d . 4, ➔ C

(G) Assertion-Reason 'fype Questions

1. ➔d 2. ➔a 3. ➔c 4. ➔d 5. ➔d.

□

5. 44800