SOLUTIONS3.

t SYNOPSIS ) PARTAA solution is a homogeneous molecular mixture of two or more substances, whose composition can be varied with in limits. Formation of solution is a physical process . A component present in lesser quantity in binary solution is referred to as solute. The absorption of Hydrogen by palladium is called occlusion. A solution of a solid in another solid is known as a solid solution. 4. Many alloys are solid solutions. 5. Copper and Zinc, Copper and Silver, Copper and Gold, Copper and Nickel, Iron and Carbon etc., form solid solutions. 6. Aqueous solutions are those prepared using water j Alcoholic solutions have alcohol as solvent Non aqueous solutions have other solvents. 7. An alloy of a metal with mercury is called an amalgam. 8. Gold amalgam is a solution of mercury in the metal. 9. Iron does not form an amalgam. Hence mercury is stored in Iron containers. 10. Smoke, fog, mist, milk etc., are not true solutions. They are called colloidal solutions. Milk is an emulsion of fatty matter in water. The components do not loose their identity during the formation of a solution. If solute is non volatile and soluent is volatile liquid, then they can be separated from the solution by evaparation or by distillation. A solution whose molar concentration is definitely known is referred as a standard solution. A solution containing relatively small quantity of solute compared to that of solvent is known as a dilute solution.11

12. 13. 14. 15. 16. The most ideal method of expressing concentration is the molality (m). 17. The commonly used method of expressing concentration is molarity (M). 18. A standard solution is prepared by using a standard flask or volumetic flask 19. Percentage by weight indicates the weight of the solute in 100 parts by weight of the solution. If W grams of a solution contains X grams of the solute, then percentage by XxlOO weight =---------20. Percentage by volume indicates the volume of the solute in 100 volumes of the solution. Percentage by volume is used only when both the solute and solvent are liquids or gases. 21 Molarity indicates the number of moles of the solute dissolved in one litre of the solution. 22. Molarity also indicates the number of millimoles of the solute dissolved in one millilitre of the solution. 23. 1 mole = 1000 millimoles 24. Number of millimoles of the solute present in V ml. of the solution = Volume X Molarity = VXM 25. Number of moles of the solute present in Vml. of the solution =-------1000 Number of moles in V litres of solution=M V 26. When a solution is diluted its molarity decreases. VM = VM

11

V

21 L2

Vj = Volume of the solution before dilution. Mj = Molarity of the solution before dilution. V2 = Volume of the solution after dilution. M2 = Molarity of the solution after dilution. (VfVj) = volume of water added V, = (M-M2) V. =(M-M2)jEAMCET - Vol - II - E.MESRICHAITANYA

EAMCET CHEMISTRY 1000 SOLUTIONS 27. 28. 29. 30. 31. 32. 33. Molarity = xm v g = weight of the solute v = volume of the solution in millilitres m = molecular weight of the solute M=IQx

GMW (x = %w/v) M=~ 10%d 34. 35. 36. GMW where %= percentage by weight d= Density in grams/ml Molarity changes with temperature. Equivalent weight of a substance expressed in grams is known as gram-equivalent weight or gram-equivalent or equivalent. Number of gram-equivalents _ Weight of the substance in grams ,rt Equivalent weight Normality indicates the number of gi^m-' equivalents of solute present in onp liltp of the solution. Normality also indicates the number of milli equivalents of solute present in one millilitre of the solution. Number of milliequivalents of solute present in V ni.1. of solution = Volume (V) X Normality (N) No. of gram equivalents of solute in V litres of solution = NV A normal solution means one Normal solution.

A decinormal solution is N/10 or 0.1N solution. A centinormal solution is N/100 or 0.01 N solution; 37. Normality = gxlOOO VxE where g = weight of the solute V = volume of the solution in ml. E = Equivalent weight of the solute.lOx

N=GEW (x = %w/v) 38. 39. 40. 41. 42. 43. N= 10%d GEW where %= percentage by weight d= Density in grams/ml When a solution is diluted, its normality decreases. V .N ii V .N2 2

(Before dilution) (After dilution) Volume of water to be added = V-V = (N-N2) x V/N2 = (NrN2) x V2/N, Normality also depends upon temperature. Fonnula weight of a substance expressed in grams: is known as gram-formula weight. Formality of a solution (F) indicates the number of gram-formula weights of the solute present in one litre of the solution.gxlOOO

Formality =T~TT where g = weight of the solute V = volume of the solute in m.l. F = formula weight of the solute. Molarity, Normality and Formality decrease with increase in temperature. Equivalent weight of Acid : E =Acid

Formula weight Basicity Number of replaceble hydrogens of acid is called Basicity. J HC1 M1 36.5 1 = 36.5Efc

-M = 98 = 49H2S042

2

P

_M98

H3PO4=7"7-32-6 HNCL } \ M 126 EH2C204.2H20 - 2 "E

2

SRSCHAITANYA

EAMCET - Vol -11 - E.M=

sm\ SOLUTIONS -p'-CHjCOOH

60El w I H3

PO:, ~ "."-,"

E M 2 M 1 =HjC204

='

2

44. Equivalent weight of Base : __ Forrnula weight of B ase w " Acidity of Base Number of replaceble hydroxyl groups of base is called Acidity. E NaOH E M 40 Mn

1 1 _M_56 KOH" i ~ f"56M 171.33 4 B(OH)2 = 85,67 C Fe(OH)3 3 _ 3 ~"3X/ 45. Equivalent weight of salt : _ Formula weight of thesalt p a S lt Total charge of the cation or Anion of the s alt F 58.5 NaCl i i ~ J*-J F 106 ___-___------- EAMCET CHEMISTRY Change in oxidation number = 7-2 = 5 or No. of electrons gained = 5 = 31.6 b) F _ 158 KMnO, ~ 5 ~ ~l~ In dil-alkaline Medium or in neutral medium: Mn04" + 3e- > MnQ2 (+7) (+4) Change in Oxidation Number = 7-4 =3 (or) Number of electrons gained = 3 F158 n KMnO, ~3~ 3 ~3Z"0 2. Potassium dichromate = K2Cr207

Cr?07" + 6e~ -> Cr+3 Change in oxidation number for one :'Cr' atom = 3 Change in oxidation number lor two 'Cr' atom = 6 Number of electrons gained = 6 F 294 K Cr 2 27 ^ 6 = 49 M=F - - ~~~~ - SI

Na,CO 2 2 fc CaC03^2~ 2 46. Equivalent weight of oxidising agent : _ Formula weight of oxidant E . =---------------------------------o an Electrons gained by oxidant Examples : 1. KMn04 = Potassium permanganate a) In acid medium : Mn04" +5e- > Mn2+ +7 +2

~Wy====^^E~^===:=~=EAMCET - Vol - II - E.M47. Equivalent weight of reducing agent : Formula weight of reductantreductant

Electrons lost by reductant Mohr's salt is Ferrous Ammonium sulphate Formula - FeS04 (NH4)2 S04. 6H20 Formula weight = 392 F J 392 1" Fe-e

-> Fe+^M ohr's salt

1 = 392 48. a) Molaritv of the mixture when two solutions of same solute are mixed V]+V2 b) Normality of the mixture when two solutions of same solute are mixed N= NtVt+ISLV. ~ V,+V,2

' c) When Vaml of an strong acid of normality Na is mixed with Vb ml of a strong base of normality Nb, i) If N V = N, V, , the solution is neutrala a b b

SRICHAITANYA /'

EAMCET CHEMISTRY

..

,

;,

ii) If N V > N.V., the solution is acidic NVa-N,,Vb [H+] its normality = V+V iii)If N V < NV., the solutions is basic NbVb-NaVa [OH] its normality = V +Vb d) In volumetric analysis, M1V1 _ M2V ii) N1V1 = N2V2 e) Normality = Molarity x Basicity of acid Normality = Molarity x Acidity of base Normality = Molarity x total +ve or -ve charge of the salt Normality = Molarity x total change in ox. state per mole of oxidant or reductantwt of base wt of acid NaVa(ml) *) GEWofbase" 1000 NbVb (ml) GEWofacid" 1000

49. Molality (m) indicates the number of mol^s of the solute dissolved in 1000 grams or one kilogram of the solvent. Molality can also be defined a-> the number of millimoles of the solute dissolved in one gram of the solvent. gXlOOO 50. Molality = ~^~ Where g = weight of the solute in grams W = weight of the solvent in grams m = molecular weight of the solute 51. Mole fraction of the solute No. of molesof the solute ~ No. of molesof solute + No. of molesof solvent Mole fraction of the solvent No. of molesof the solvent_____ No. of moles of solute+ No. of moles of solvent 52. Mole fraction of Solute + Mole fraction of Solvent = 1 53. Molality and mole fraction are independent of temperature. ________________________ SOLUTIONS 54. Relationship between molarity and normality is Normality X Equivalent weight = Molarity X Molecular weight (or) MolarityX Molecular Weight Equivalent Weight Normality = 55. Problem: Concentrated sulphuric acid contains 98% sulphuric acid. Its specific gravity is 1.8 What are the Molarity and Normality of concentrated sulphuric acid Solution: a) 98% sulphuric acid means 100 g. of the solution contains 98 g. of H2S04 For calculating molarity we require the volume of the solution. Volume of 100 grams of the solution = Mass 100 , density 1.8 Molarity = Molarity = gxlOOO VXrn 98x1000 1000x1.8 -18 (100/1.8)x98 1000 General formula for calculating molarity from percentage is Molarity = Weight pe-rcentagex densityx 10 Molecular Weight

Normality = Moiarityx Molecul ar Weight Equivalent Weight 18x98 49 = 18 x 2 = 36 General formula for calculating normality from percentage is Normality = Weight Percentagex density x 10 Equivalent Weight 56. Problem:The density of 1 M solution of Sodium hydroxide is 1.02 gm/ml. What is its molality ?SRiCHAITANYA

: EAMCET - Vol - II - E.M|

SOLUTIONS ----------------________________ Solution: 1 M solution means 1000 ml. of the solution contains 1 mole or 40 grams of NaOH. Mass of 1000 m.l. of the solution = Volume x density = 1000 X 1.02 = 1020 grams 1020 grams of the solution contains 40 grams of NaOH. Mass of water = 1020 - 40 = 980 grams. . Molality= gXl0Q0 = 40xl00 -10Q0 _1QQ Wxm 980x40 980 = 1.022 molal 57. Problem: What is the Normality of 0.2M H2S04.-, , ..J

98

r

..

MolarityxMolecularWeight

Solution: Normality--------r ---. EquivalentWeight

-

59. 0.2x98 49 = 0.2X2 = 0.4 58. Problem: The density of a 1 m solution of NaOH is 1.02 gm/ml. What is its molarity? Solution: For calculating the molarity we require the volume of the solution. lm solution means 1 mole of NaOHf(4;CH. grams) is dissolved in 1000 grams of water. Mass of the solution = 1000 + 40 = 1040 grams. Volume of the solution gxlOOO Mlarity = "V^rT Mass 1040 density 1.02 40x1000

mi.(1040/1 l.Q2)x40 1000x11.02 1020 102 104 51 = 0.98M 1040 1040 104 52 The molality of a solution is greater than its molarity. When Vj C.C. of one solution completely reacts with V2 C.C of another solution, then the number of milliequivalents of the two solutes in the solutions should be equal. Therefore VrN1 = V2.N2 where Nt and N7 are the normalities of the two solutions respectively. If molarities are used, then

HOEAMCET CHEMISTRY 60. 61. 62.v M

r i_v2-M2

"1 "2 where n: and n2 are the number of" moles of the two reactants participating in the reaction. Problem : Concentrated H,SQ4 is 36 N. What is the volume of concentrated H2S04 required to prepare 500 m.l. of N H2S04 ? Solution : Vr Nj = V,. N, VJ x 36 = 500 x 1 .-. V = = 13.89ml. 1 36 Problem : What is the mass of NaOH required to completely neutralise 100 ml. of 0.2N HC1? Solution: 100 ml. of 0.2N HC1 s 100 ml. of 0.2 N NaOH. Hence we have to find the mass of NaOH present in 300 ml. of 0.2 N NaOH Normality gxlOOO VxE or 0.2 2X1000 100x40 Mass = 0.2x100x40 1000 = 0.8 gram. Problem : 100 m.l. of 0.2 N HC1 is completely neutralised by 0.8 gram of a diacid base. What is the molecular weight of the basic? Solution: Let 0.8 grams of the base be present in 100 ml. of the solution. Then the normality of the base solution will also be 0.2. N= gxlOOO VxE /. 0.2: 0.8x1000 E= 100XE 0.8x1000 = 40 100x0.2 Since the acidity of the base is 2, Molecular weight = 2 x Eq.wt =2x40 = 80 63. Problem: 100 ml. of 0.2 N H2S04 is mixed with 100 m.l. of 0.1 N NaOH. What is the normality of the resultant solution ? EAMCET - Vol II - E.M: SRI CHAITANYA EAMCET CHEMISTRY Solution : General formula N3 = 2. 3.v N v

r r 2;N2 VV2J

64. No. of milli equivalents of H2S04 taken = V1.N1 = 100 x 0.2 = 20 No. of milli equivalents of NaOH taken = V2.N2= 100x0.1 = 10 No. of milliequivalents of H2S04 remaining unreacted = 20 - 10 = 10 No. of milli equivalents of solute n ^ ' Volumeofthesolutioninml.=

i2.=_L=0.05

200 20 Problem: 100 m.l. of 0.1 N HC1 is mixed with 100 ml of 0.1 N HN03. What is the normality of the resultant mixture with respect to acid? Solution: Here both solutions are acid solutions. There is no reaction between the two.N,

WV2;N2

ivv2J100x0.1+100x0.1 200 = 0.1N,

EXERCISE - 1If air is taken as a binary solution, the solvent is 1) N, 2) 07 3) C02 4) R> Homogeneous system among the following is 1) milk 2) sand in water 3) urea in water 4) benzene in water The physical change among the following is 1) burning of coal 2) burning of sulphur 3) dissolution of Glucose in water 4) burning of white phosphorus A binary solution constitute......number of phases 1) one 2) two 3) three 4) four A mixture of salt and water can be separated by 1) filtraction 2) decantation 3) crystallisation 4) kept long standing __________________-----------SOLUTIONS 6. 100 grams of an aqueous solution contains 10 grams of Glucose. Then solute and solvent are 1) Water and Glucose 2) Glucose and Water 3) Alcohol and Glucose 4) Glucose and Alcohol 7. Which of the following differ from the others 1) ruby 2) elektron 3) bell metal 4) amalgam 8. Occlusion of Hydrogen on Palladium is an example for.....type solution 1) gas in solid 2) solid in gas 3) gas in liquid 4) liquid in gas

9. The ratio of the number of moles of solute to the total number of moles of solute and solvent is kriown as 1) Molarity 2) motelity 3) mole fraction of solute ,, 4|mole fraction of solvent iO. The apparatus used to prepare "a standard solution" is 1) standard flask or volumetric flask 2) burette 3) pipette 4) conical flask 11. Which of the following represents Imolar (1M) solution 1) 1 gram of solute in 1 litre of the solution 2) 1 gram of solute in 1 kg of the solution 3) 1 gram mole of solute in 1 kg of solvent 4) 1 mole of solute in 1 litre of the solution 12. The units of Molarity are 1) gms. lit"[ 2) moles, lit" 3) equivalents, lit-1 4) moles, kg' 13. Molarity of a solution 1) increases with increase of temperature 2) decreases with increase of temperature 3) may increase or decrease with increase of temperature 4) is independent of temperature 14. The number of gram moles of solute present per litre of the solution is known as 1) Molarity 2) Molality 3) Normality 4) Mole fraction1

-l

SRICHAITANYA

iEAMCET - Vol - II - E.M| 25. 26. 27. 28. 29. SOLUTIONS ________________________ 15. A solution whose concentration is known is called 1) a saturated solution 2) a colloidal solution 3) a standard solution 4) an unsaturated solution 16. To halve the molarity of a solution the following should be adopted 1) weight of the solute to be doubled 2) weight of the solvent to be doubled 3) volume of the solvent to be doubled 4) volume of the solution to be doubled 17. 100 milli moles of a solute is present in 200 ml of the solution. Then its molarity is 1) 2M 2) 1M 3) 0.5 M 4) 1.5M 18. 0.5 moles of a solute is present in 500 ml of the solution. Then its Molarity is 1)1M 2)10_3M _2 3)10 M 4)10_1M 19. The number of milli moles of solute present in 10 ml of decimolar solution is

1) 1 2) 10~3 2 3) 104) 10_1 20. The number of moles of solute present In 10 ml of decimolar solution is 1) 1 2) 10"3 -2 3) 10 4) 10_i 21. The number of moles of solute present in 0.5 dm3 of 0.5M solution is 1) 0.5 2) 5 X 10~3 2 3) 5 X 10~ 4) 0.25 22. 4 g. of NaOH is present in 100 ml solution. Its molarity is 1) 1M 2) 2M 3) O'.IM 4) 0.2M 23. 2 gms of NaOH is present in 1 litre of the solution, The Molarity of the solution is D0.5M 2) 0.05 M 3) 0.1 M 4) 0.005 M 24. 4.9 g of H2S04 is present in 500 ml of the solution. The Molarity of the solution is DO. 1M 2) 0.2 M 3)0.02 x 10~2M 4) 0.05M =EAMCET - Vol - II 30. 31. 32. 33. 34. 35. ----------------------- EAMCET CHEMISTRY 200 ml of a solution of Barium Hydroxide contains 171.4 mg of solute. The Molarity of the solution will be [Mol.wt. of Ba(OH)2 = 171.4] 1) 5 x 10~2 M 2) 0.5 M 3 3) 5 x 10~ M 4) 5 x 10~4 M The weight of AgNO, (Mol.wt = 170) present in 100 ml of its 0.25M solution is 1) 4.25 g 2) 42.5 g 3) 8.5 g 4) 2.125 g The weight of H2S04 present in 400 ml of 0.125M solution is 1) 2.45 g 3) 4.9 g 2) 3.92 g 4) 9.8 g The Molarity of 2.925% (W/V) sodium chloride solution is 1) 0.5 M 2) 1.0 M 3) 5.0; M 4) 0.25 M of 1% (W/V) NaOH The Molarity solution is -1) 0.5 M 3) 0.125 M 2) 0.25 M 4) 1.0 M The Molarity of 4.9%(W/V) K2Cr20? (Mol.wt = 294) solution is 1) 1.66M 2) 0.332 M 3) 0.166 M 4) 3.32 M The number of moles of KCl present in 500 ml of 2M solution is 1) 1 2) 2 3) 0.5 4) 4 The number of moles of solute present in 250 ml of 0.01 M solution is 1) 2.5 x 10 3) 2.5 x 10"-l

2) 2.5 x 10 4) 2.5 x 10-2 -4

2.12 g. of anhydrous Na2C03 present in how much volume of the solution will have 0.2 M concentration 1) 200 ml 2) 100 ml 3) Hit 4) 0.5 lit 5 millimoles of solute present in how many litres of the solution will have 0.1 M 1) 5 2) 0.5 3) 50 4) 0.05 3.15 grams of solute is present in 200ml of 0.25 M solution. The solute may be 2) HC1 4) HNCL 1) H2C204 . 2H20 3) H2S04 SRI CHAITANYA EAMCET CHEMISTRY 36 Cone. H,SO, is 18M. The volume of2 4

conc.tLSO. required to prepare 250 ml of 4M H SO. solution is 37. ..... ml 1) 55.5 2) 194.5 3) 305.5 4) 19.45 The volume of water to be added to 100ml of 0.5M iNaCl solution in order to make it decimolar is 1) 500ml 2) 400ml 3) 600 ml 4) 50 ml The volume of water to be added to convert 10 ml of deca molar HC1 solution to deci molar solution is 1) 99 ml 2) 100 ml 3) 1000ml 4) 990 ml 39. NaOH solution is labelled as 10 % by volume. Then the molarity of NaOH solution is 1) 2 M 2) 2.5 M 3) 4M 4) 1M 38. 40. 41. H2S04 is labelled as 0.98% (w/v). Then the molarity of H,SO solution is 1) 0.1 M " 2) 0.2 M 3) 0.25 M 4) 2.5 M J;; 50 cc of decinormal NaOH solution will he completely neutralised by 'x' ml of decimolar HJSO. solution. The value of V is2 4

42. 1) 1 2) 10 In the reaction 2NaOH + H^PO,3 4

3) 25 4) 100 -> NaJHPO, + 2HO,2 4 2

the Equivalent weight of the acid is

1) 49 2) 98 3) 32.6 4) 36.5 43. The units of Normality are 1) moles, lit" 2) moles. Kg-l -l t

44. 3) equivalent, lit 4) equivalents, kg M = Molarity of a solution 00 = %(w/w) by weight of solute d = density of the solution (in g. ml-1) m = Gram mol. wt of solute Then the correct relationship is-l

,-h Y)M = (uxd 3)M=m (uxlOxd 2)M = mxd 4) M =0)

coxlOOOxd mSRI CHAITANYA

m : EAMCET _--------------------------------_ SOLUTIONS 45. The following is not a fixed quantity 1) atomic weight of an element 2) equivalent weight of an element (or) compound 3) molecular weight of a compound 4) formula weight of a substance 46. In the reaction HXO, + NaOH > NaHCO, + H_0, the2 3 3 2'

equivalent weight of carbonic add is 1) 31 2) 62 3) 20.6 4) 124 47. The equivalent weight of CuSO when it is converted to Cu I 1) 3V M1 M M2)

T

4)2M

48. Which of the following acid has the same molecular weight and equivalent weight 1) H?P02 2) H3P03 1)H3P04 2)H2S04 49. The equivalent weight of Hypo in the reaction [M = molecular weight] 2Na2S203 + I2 -M -> 2NaI + Na,S .O, is2 4 6

1) M 2)M3) 4)

T T4 2

M 50. The equivalent weight of H2S04 when it reacts according to the equation H SO, + NaOH2 4

-NaHS04 + HO 1) 49 2) 98 3) 32.6 4) 196 51. If 'M* is the molecular weight of CaC03, its equivalent weight is M M M1)M 2)

T

3)

T

4)

T

52. 'M' is the molecular weight of KMn04The equivalent weight of KMn04 when it reacts according to the equation 2KMnO, + 3H,SO, + SHX^O, >4 2 4 2 4 4 2 2 4 2 2

JCSO^ + 2MnSO + 8H.O + 10CO. 1) M Vol - II 2 E.M= 2)M 3)M 4)M 93SOLUTIONS -----------------------------------

53. Molecular weight of H3P04 is 'M*. Its equivalent weight when it reacts according to the equation 94 -> CaHPO, + 2H,04 2

H3P04 + Ca(OH)2 M M 1)M 2)

3)y

4)3M

54.4

Molecular weight of KMnO is "M". In a reaction KMnO. is reduced to KJVlnO,.2 4

The equivalent weight of KMnO is M M M DM 2)T 3)y 4)T 55. The equivalent weight of KMnO when it reacts according to the equation Mn04 + 2H20 + 3e" -> Mn02 + 40H1)M2)'

M3)-

M4)'

M 56. M is the molecular weight of FeSO,. Its equivalent weight when it reacts according to the equation 2KMn04 + 10FeSO4 + 8H2S04 -^M K.SO, + 2MnSO, + 5Fe,(SOJ, + 8H,62 4 4 2 43 ^ 2

M

DM 2)y2:

M 3)

4VM 5 7. The equivalent weight of Iodine in the reaction;ii

M 2Na.S-O, + I, > 2NaI + Na.S.0- is2 2 3 2 2 4 6

1)M M2

>v

3)'

4) 2M 58. Equivalent weight of K2Cr20? in acidic medium is M M

"73) M M4)

2)

T

T

5 9. The equivalent weight of Hypo in the reaction Na2S203 + Cl2 + H20 -> Na2S04 + 2HC/ + S is M M DM 2)~ 3) 2 3 4)2M 60. Molecular weight of Mohr's salt is 392. Its equivalent weight when it is oxidised by

KMnO in acidic medium is 1) 392-------------------------------- EAMCET CHEMISTRY

61.4

The equivalent weight of CH. in the reaction

63. 65.66,

67. 68. 69.70.

2) 196 3) 130.6 4) 78.5 EAMCET - Vol - L CH, + 20,4 2

- C02 + 2H20 is

DM2)

M3)

M 124)

M 16 62. The equivalent weight of Glucose in the reaction C H 6 126 + 62 ^ 6C02 + 6H20 is M

"7M2)

TI

3)

M 244)

M 48 10 milli equivalents of solute is present in 5 ml of an aqueous solution. Then its Normality is 1) 1 N 2) 0.5 N 3) 2 N 4) 0.25 N 64. 10" gram equivalent weights of solute is present in 10 ml of solution. Then its normality is 1) 0.1 N 2) 1 N 3) 0.01 N 4) 0.01 N The number of milli equivalents of solute present 10 ml of 0.1N solution are 1)1 2) 10 3) 100 4) 0.1 250ml of a solution of Na CO contains 4.24g of solute. Its Normality is 1) 0.16 N 2) 0.32 N 3) 0.64 N 4) 0.8 N 3.7g of calcium hydroxide is present in 100 ml of solution. The Normality of the solution is

1) 2 N 2) 0.5 N 3) 1 N 4) 0.25 N 9.8 g of Orthophosphoric acid is dissolved in water and the solution is made upto 2 litres with distilled water. The Normality of the solution is 1) 0.75 N 2) 0.05 N 3) 0.3 N 4) 0.15 N The number of equivalents of HC1 in 200 ml of 0.1N HC1 solution is 1) 2 2) 0.23) 0.02M

4) 20

The weight of crystalline Oxalic acid (H2C204. 2H,0) required to prepare 100 ml of 0.05N solution is 1) 0.1575 g 2) 1.575g 3) 0.315 g 4) 0.63 g E.M=SRI CHAITANYA EAMCET CHEMISTRY------------------------------

71. The volume of water to beadded to 10ml of ION HO solution to make it decinormal is 1) 1000ml 2) 990 ml 3) 99 ml 4) 9,990 ml 72. Cone. HC1 is 12N. The volume of Cone. HC1 required to prepare 2 litres of 4N solution is 1) 1333.4 ml 2) 2666.6 ml 3) 1666.6 ml 4) 666.6 ml 73. 40 ml of 0.2N H2S04will be neutralised by 20 ml of......... NaOH solution 1) 0.4 N 2) 0.1 N 3) 0.8 N 4) 0.2 N 74. 20 ml of 0.1 N FeS04 solution will be completely oxidised by____ml of 0.05N KMnO solution in acidic medium 1) 20ml 2) 10ml 3) 40 ml 4) 80 ml 75. The Normality of 0.98% (w/v) H2S04 solution is 1) 0.1N 2) 0.2N 3) 0.4N 4) IN 76. H2S04 # solution is labelled as 0.98% by weight. Specific gravity of the solution is 1.08. Normality of the solution is ' 1) 0.108 N 2) 0.216 N 3) 0.98 N 4) 0.049 N 77. Molarity of 3N H3P04 solution is 1) 9 M 2) 1.5 M 3) 6 M 4) 1 M 78. The volume of 0.05N Nl2C03 solution required to neutralise 200ml of 0.02M H SO solution is 1) 80ml 2) 140ail 3) 160ml 4) 240ml 79. Molecular weight of a dibasic acid is 'M\ If 'V ml of the solution contains ' 0)' gms of the acid, the Normality of the solution is_____________________ SOLUTIONS

81. To change the molal cone, to one half, one of the following should be adopted 1) weight of the solute should be doubled

2) weight of the solvent should be doubled 3) volume of the solvent should be doubled 4) weight of the solution should be doubled 82. 6 g. of Urea is present in 100 g. of water, the concentration of the solution is 1) 1M 2) 1m 3) IN 4) IF 83. A one molal solution is one that contains 1) 1 g. of the solute in 1000 g. of solvent 2) 1 g. moleof solute in 1000 ml of solution 3) 1 g. mole of solute in 22.4 lits of sol ution 4) 1 g. mole of solute in 1000 g. of solvent 84. 4 g. of NaOH is dissolved in 1kg of water. The concentration of the solution is best expressed as 1) 0.1M 2) 0.1N 3)0.1m 4)0. IF 85. 0.1 gram mole of urea is dissolved in lOOg. /' of water The molality of the solution is 1) 1 m 2) 0.01 M 3) 0.01 m 4) 1.0 M 86. The molality of 2% (W/W) NaCl solution nearly 1) 0.02m 2) 0.35 m 3) 0.25 m 4) 0.45 m 87. The molality of 10 % (WAV) NaOH solution is 1) 2.77 m 2) 5.54 ra 3) 0.0025m 4) 2.5 in 88. The concentration unit which is 1)3)'

2 x cox 1000 MxV coxlOOO2) 2xco

MxV4)(si

2xMxV 2xMxV 80. At 25C for a given solution M = m, then at 50C the correct relationship is l)M = m 2)M>m 3)M.2::

9)314)2

llil!24)3 29M

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EXERCISE - 6On dissolving 1 mole of each of the following acids, in 1 litre water, the acid which does not give a solution of strenght IN (MPMT) 1) Hydrochloric acid 2) Perchloric acid 3) Nitric acid 4) Phosphoric acid When 10 ml of 10 M solution of HjS04 and 100 ml of 1 M solution of NaOH are mixed the resulting solution willbe (JIPMER) 1) Acidic 2) Neutral 3) Weakly alkaline 4) Strongly alkaline A 0.01 M ammonia solution is 5% ionised. The concentration of OH" ions is (Karanataka EEE) 1) 5xlO-3 M 2) 5xl0-*M 3) lxlO^M 4) 5xlQ-2 M jSRICHAITANYA iEAMCET - Vol - II - E.M= SOLUTIONS -- --------_______________ 4. The density of NH, OH solution is found to be 0.6 g/ml. It contains 34% by weight of NH4OH, Calculate the normality of the solution. (Kerala, EEE) 1) 4.8 N 2) ION 3) 0.5 N 4) 5.8 N 5. The vapour pressure of a solution of 5 g of non electrolyte in 100 g of water at a particular temperature is 2985 N/rn2. The vapour pressure of pure water is 3000 N/ m2. The molecular weight of the solute is

(IIT) 1) 180 2) 120 3) 60 4) 392 6. If 5.85 g of NaCl is dissolved in 90 g of water, the mole fraction of NaCS is (MP) 1)0.1 2)0.01 3)0.2 4)0.0196 7. How many grams of CH3OH would have to be added to water to prepare 150 ml of a solution that is 2 M CH3OH? (CBSE) 1) 9.6 2) 2.4 M 3) 9.6X104 4) 4.3xl02 8. 10 mi of cone. H2S04 ( 18 molar) is dtiuteel to 1 litre, the approximate strength .of dilute acid could be ( JIPMER) 1) 0.18 N 2) 0.09 N 3) 0.36 N 4) 18 N 9. 100 ml of 0.3 N HCI is mixed with 200 ml of 0.6 N HLS04. The final normality of the resulting solution will be ( Delhi PMT) 1) 0.1 N 2) 0.2 N) 3) 0.3 N 4) 0.5 N 10. The normality of the acid solution obtained by mixing 100 ml of N/2 H2S04 solution and 300 ml of N/10 HCI solution is (AIIMS) 1) 0.2 2) 0.5 3) 0.6 4) 0.3 11. 1.5 litres of a solution of normality x and 2.5 litres of 2 M HCI are mixed together. The resultants solution had a normality 5. Then the value of x is (NSTS) 1) 6 2) 10 3) 8 4) 4 12. The vapour pressure of pure benzene is 640 mm at 298 K. A solution of a solute in benzene shows a vapour pressure of 630 mm at the same temperature. Then the mole fraction of the solute is (NSTS) 1) 0.016 2) 0.18 3) 0.20 4) 0.25 -----------.------ EAMGET CHEMISTRY 13. A sample of Na2CO,.H20 weighing 0.62 is added to 100 ml of 0.1N H,S04 solution. What will be the resulting solution? (Manipai) 1) Acidic 2) Neutral 3) Basic 4) Both 1 and 3 14. The vapour pressure of a solvent decreased by 10 mm of mercury, when a non-volalite solute was added to the solvent. The moire fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is to be 20 mm of mercury? ( CBSE) 1)0.2 2)0.4 3) 0.6 4) 0.8 15. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. a nonvolatile and non-electrolyte solid, weighing 2.175 grn is added to 39.08 gm benzene. The vapour pressure of the solution is 600 mm f Ji of Hg. What is the molecular weight of solid substance? (CBSE) 1) 79.8 2) 69.5 3) 65.25 4) 59.6 16. How many grams of dibasic acid (mol. weigh 200) should be present in 100 ml of the aqueous solution to give a strength of 01 N, (CBSE) 1) 10 g 2) 20 g 3) 1 g 4) 2 g 17. Concentrated sulphuric acid in a bottle is labelled as 49% by mass and density is 1.5 gms cm3. The molarity of the solution is (CBSE) 1) 5.7 2) 9.6 3) 7.5 4) 15 18. A sample of rectified spirit contains 46 percent of ethanol by mass, the mole fractions of

ethanol and water are (CBSE) 1) 0.80 and 0.20 respectively 2) 0.20 and 0.80 respectively 3) 0.75 and 0.25 respectively 4) 0.25 and 0.75 respectively 19. The number of moles of KMnO that will4

be needed to react completely with one mole of ferrous oxalate in acidic solution is (IIT) 1) 3/5 2) 2/5 3) 4/5 4) 1

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EAMCET CHEMISTRY------------------------20. The number of moles of KMn04 that will be needed to react with one mole of sulphite ion in acidic solution is (HT) 10) 2/5 2) 3/5 3) 4/5 4) 1 21. The molarity and molaiity of a solution of sulphuric acid are 11.07 and 21.91 respectively. The density of the solution in gm/ml is (CBSE) 1)4.36 2) 0.795 3) 2.18 4) 1.59 22. The normality of a mixture obtained by mixing 50 ml of 0.01 N HC1 and 25 ml of 0.01 N NaOH is (Manipal) 1) 0.025 2) 0.0066 3) 0.125 4) Zero 23. The normality of 0.3 M phosphorus acid (H3P03) is (HT)

1) 0.1

2) 0.8

3) 0.3 4) 0.6 .| rf 24. The number of milliequivalents in 100 mlio|l 0.5 N HC1 solution is (A I IMS) 1) 200 2) 100 3) 50 4) 25 '...25. The volume of water to be fidd 'to 100 cm3 of 0.5 N H2S04 to ge^decinormal concentration (Karnataka CEE) 1) 400 cm3 2) 450 cm3 3) 100 cm3 4) 500 cm3 26. An aqueous solution of 6.3 g oxalic acid di-hydrate is made upto 250 ml. the volume of 0.1 N NaOH required to completely neutralise 10 ml of this solution is (HT) 1) 40 ml 2) 20 ml 3) 10 ml 4) 4 ml 27. 100 ml of 0.3 M HNO, and 200 ml of 0.3 M HjS04 are mixed. The normality of resulting solution is (NSTS) 1) 0.9 2) 0.6 3) 0.4 4) 0.5 28. the volume of 0.2 N base required to completely react with 0.5 litre of 0.1 N acid is (CBSE) 1) 1 litre 2) 0.1 litre 3) 0.25 litre 4) 0.01 litre

__________________------ SOLUTIONS

29. 100 ml of 0.3 N HC1 is mixed with 200 ml of 0.6 N H2S04. The final normality of resulting solution will be (Delhi PMT) 1) 0.1 N 2) 0 2 N 3) 0.3 N 4) 0.5 N 30. 1.25 g of solid dibasic acid is compltely neutralised by 25 ml of 0.25 M Ba(OH)2 solution. Then the molecular mass of the acid is ' (NSTS) 1) 100 2) 150 3) 120 4) 200 31. To convert 12 g of NaH,P04 completely into Na P04, the volume of lmolar NaOH re(NSTS) 2) 100 cm3 4) 120 cm' quired is 1) 200 cm3 3) 80 cm3 32. Normality of a 2 M sulphuric acid is (A1IMS) 1)2N 2)4N 3)N/2 4) N/4 33. J^jfch of the following solutions of H,S04 will exactly neutralize 25 ml of 0.2 M NaOH? (AIIMS) -:?" 1) 12.5 ml of 0.1 M solution 2)25 ml of 0.1 M solution 3) 25 ml of 0.2 M solution 4) 50 ml of 0.2 M solution 34. The relative lowering of vapour pressure of a solution of 6 g of urea in 90 g of water is close to (NSTS) 1) 0.02 2) 0.04 3) 0.06 4) 0.03 35. Increasing the temperature of an aqueous solution will cause (IIT) 1) Molaiity to decrease 2) Mole fraction to decrease 3) Molarity to decrease 4) Percentage by weight to increase

ANSWERS1)4 11)2 16)3 4 1 31)1 2)1 7)1 12)1 17)3 22)2 27)4 32)2 3)2 8)1 13)2 18)4 23)4 28)3 33)3

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