Solutions & Colligative Properties Iit

Note: Key to the questions and updates, if any, can be downloaded from http://groups.google.com/group/adichemadi Prepared by V. Aditya vardhan [email protected]

Hint :

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1)

SOLUTIONS & COLLIGATIVE PROPERTIES IIT-JEE CONCENTRATION Match the following A) Gas in solid 1) Soda water B) Solid in liquid 2) Brine solution C) Solid in solid 3) Air D) Gas in liquid 4) H2 occluded in Pd 5) Magnalium Correct matching A B C D 1) 4 2 5 3 2) 4 5 2 1 3) 4 2 3 5 4) 4 2 5 1 Soda water

=

CO 2 in water

Brine

=

NaCl in Water

=

O 2 (and other gases) in N 2

=

Mg (1-15%) in Al (99-85%)

Air

Magnalium

2)

Not a homogeneous mixture 1) Soda water 3) Sodium amalgam

2) aqueous sugar solution 4) Milk

Note : In case of milk, fine liquid fat particles are dispersed in water. It is not a true solution.

3)

The molarity of 500 cc of solution containing 0.2 moles of NaCl. 1) 0.2 mole L-1 2) 0.4 mole L-1 3) 0.1 mole L-1

4) 0.8 mole L-1

Hint : Molarity is no. of moles of solute per litre of solution.

4)

The molarity of H2SO4 solution is 2 M at 27oC. What will be its molarity at 100oC ? 1) = 2 M 2) < 2 M 3) > 2 M 4) All Hint : Molarity is altered with volume of the solution.

5) 6)

The molarity of 250 ml of solution containing 0.365 g. of HCl. 1) 0.04 M 2) 0. 04 m 3) 0.01 M What is the molarity of 2% NaOH solution ? 1) 20 M 2) 10 M 3) 0.5 M 

Formula

7)

:

4) 0. 02 M

4) 0.2 M



%  WV  X 10

M

Formula weight

The molarity of 5.3% Na2CO3 solution of specific gravity 1.1 is 1) 0.55 M 2) 5.3 M 3) 0.53 M 

4) All



%  W  X 10 X d

Formula : M 

8) 9)

W

Formula weight

The weight of H2SO4 in 200 mL of 0.2 M H SO4 solution is 2 1) 3.92 grams 2) 39.2 grams 3) 9.8 grams What is the weight percentage of 1 M Na2 SO4 solution ? 1) 1.42% 2) 14.2% 3) 0.29%

4) 0.98 grams 4) 2.9%

Note : Weight percentage is weight of solute in 100 ml of solution.

10) The weight percentage of 4 M H2O2 solution of the specific gravity is 1.36 is

Prepared by V. Aditya vardhan [email protected]

1) 20 %

w w

2) 10 %

w V

3) 10 %

w w

4) None

-1

11) The density (in g. mL ) of a 3.60 M sulfuric acid solution that is 29% H2 SO4 by mass will be 1) 1.22 2) 1.45 3) 1.64 4) 1.88 (AIEEE-2007) 12) 4.6 g of ethyl alcohol is dissolved in 1000 g of water. The molality of the solution is 1) 4.6 m 2) 0.1 m 3) 0.46 m 4) 0.01 m Hint : Molality is no. of moles of solute dissolved in 1 Kg of solvent.

13) Two moles of glucose is dissolved in 500 mL of water. The molality will be 1) 1.8 m 2) 18 m 3) 10 m 4) 4 m

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Hint : Density of water is approximately 1 g. mL-1

14) The weight of NaOH in 10 mL of 2 m NaOH solution of specific gravity 1.08 is 1) 0.8g 2) 1.6 g 3) 2 g 4) 1.08 g w

15) The molality of 20% w NaOH solution is. 1) 20 m

2) 2 m

3) 6.25 m

4) 2.65 m

Hint: weight of solvent = weight of solution - weight of solute

16) What is the molality of 9.8% H2SO4 solution of specific gravity 1.098 1) 1 m 2) 1.098 m 3) 9.8 m 4) 2 m Hint : First calculate weight of the solution weight = density X volume

17) 120 g of NaOH is added with one litre of water. The molarity of the solution formed is 1) 3 M 2) > 3 M 3) < 3 M 4) None Hint : The final volume of the solution is more than one litre

18) The molarity of pure water is 1) 5.55M 2) 55.5M 3) 1M 4) 100M 19) For a given aqueous solution , the relation between molality (m) and molarity (M) is 1) m = M 2) m < M 3) m > M 4) All 20) One mole of a solute, occupying 100mL, is dissolved in 1000 g of a liquid ‘A’. The density of liquid ‘A’ is 2 g.mL-1 If the solution formed is ideal, then 1) M = m 2) M > m 3) M < m 4) None 21) 20 g of NaOH is present in one litre of solution. Its normality is 1) 0.5 N 2) 1 N 3) 2 N 4) 20 N Formula : Normality ( N ) 

W 1 x G. E.W V (in lit )

22) 4.9 g of H2SO4 is present in 250 mL of solution. The molarity and normality of the solution are , respectively. 1) 0.1 M & 0.2N 2) 0.2M & 0.4N 3) 0.2 M & 0.2N 4) 0.1 M & 0.1N 23) The weight of H3PO4 in 500 mL of 0.3N H3PO4 solution is. 1) 0.3 g 2) 9.8 g 3) 0.98 g 4) 4.9 g Hint : Basicity of H PO = 3 3 4

24) The normality of 5.7% Al2(SO4)3 solution is 1) 5.7 N 2) 2 N 3) 1 N Hint: EW of Al 2 (SO 4 ) 3 

4) None

FW 6

25) The specific gravity of 31.6% w/w KMnO4 solution is 1.5. The normality of the solution in the acidic medium is 1) 15 2) 1.5 3) 31.6 4) 3.16

Prepared by V. Aditya vardhan [email protected]  w %  X d X 10  w Formula : N  G .E .W

Hint :

Mn

7+

 Mn

2

(in acidic medium)

26) The normality of 2.36M H2SO4 solution is 1) 4.72N 2) 1.5 Formula : N = M X basicity

3) 1.18N

4) None

3) 0.1M

4) 0.15M

(for acids)

27) The molarity of 0.3N Ba(OH)2 solution is 1) 0.6M 2) 0.3M Formula : N = M X Acidity

(for bases)

Hint :

Mn

7+

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28) 5.26 g of KMnO4 is present in 100 cc of a solution. What is its normality in basic medium. 1) 5.26N 2) 1N 3) 3.16N 4) 2N  Mn

4

(in basic medium)

29) The number of H+ ions in 100 mL of 0.1M H2SO4

1) 6.023 X 1021 2) 1.2046 X 1022 3) 3.10 X 1021 4) All 30) The solution with least molarity is 1) 1N HCl 2) 1N H2SO4 3) 1N Na3PO4 4) None 31) The amount of HNO3 to be added to double the concentration of 500mL of 1M HNO3. 1) 63 gr 2) 6.3 gr 3) 31.5 gr 4) 12.6 gr 32) The ratio of volumes of 2 M and 4M solutions to be mixed to get 500mL of 3M solutions. 1) 1 : 2 2) 1 : 1 3) 2 : 1 4) 2 : 4 33) 500 ml of 0.4M Na3PO4 solution is diluted to 0.1M solution. The volume of solvent required is. 1) 2000 mL 2) 1500mL 3) 1000mL 4) All H in t :

U se

M V  M V 1 1 2 2

34) 300 mL of 2M NaOH solution is added to 200 mL of 0.5M NaOH solution. What is the final molarity? 1) 14M 2) 1.4M 3) 0.7M 4) 7M Hint: M mix 

M 1V1  M 2V2 V1  V2

N N HCl solution is mixed with 125 mL of H 2 SO4 and the resultant solution is 2 5 made up to 1 Litre. The normality of the final solution is 1) 0.03 N 2)0.01 N 3) 0.3 N 4) 0.5 N 36) What are the volumes of 3M HCl and 5M HCl solutions, respectively, to be mixed to get 2 litres of 3.5M HCl solution? 1) 1.5 L & 5 L 2) 2 L & 3 L 3) l L & l L 4) 1.5 L & 0.5 L

35) 10 mL of

37) Equal volumes of lM KNO3 and lM Al (NO3)3 solutions are mixed. The concentration of NO3 ions in the final solution is 1) 2M 2) 1M 3) 4M 4) 0.5M

-

-

Note : 1M Al(NO 3)3 contains 3M NO ions. 3

38) 10 mL of 1M Na2 CO3 solution is made upto 1 litre by adding water. The normality of the final solution is 1) 0.01 N 2) 0.1 N 3) 0.2 N 4) 0.02 N

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39) 20 mL of 1M NaCl is added to 10 mL of 2 M AgNO3 . The concentration of Cl ions in the final solution is 1) 0.5M 2) 0. 4M 3) 1.5M 4) None -

-

Note : All the Cl ions are precipitated out as AgCl. Hence the final concentration of Cl will be almost zero.

40) Which of the following solutions contains more number of ions? 1) 9M KNO3 2) 8M NaCl 3) 4M Al2 (SO4 )3 4) 6M Na2 SO4 41) A 0.02N NaCl solution is diluted by 100 times. The number of moles of NaCl will be 1) increased 2) decreased 3) unchanged 4) Cannot say Hint : There is no change in the number of moles of solute during dilution. Only the concentration is changed.

42) Which of the following graph represents the variation of molality (m) with temperature (T)

Molality (m)

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2)

Molality (m)

1)

T emperature (T)

Molality (m)

4)

Molality (m)

3)

T emperature (T)

T emperature (T)

T emper ature (T )

Hint : Molality is independant of temperature

43) The equivalent weight of K2 Cr2 O7 , when it is used as an oxidising agent in acidic medium is (gram atomic weights of K=39.1, Cr = 52 , O = 16) 1) 98 g 2) 49 g 3) 294 g 4) None Hint: Cr6+

GEW 



Cr3+

(in acidic medium)

Formula weight 6

44) Match the following Reaction

Equivalent weight of the reactant

A) C2 O4 2-



CO2

1)

GMW 2

B) FeSO4



Fe3+

2)

GMW 1

C) H3 PO4



PO4 3-

3)

GMW 3

D) KMnO4



Mn2+

4)

GMW 5

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Correct matching is A B C D 1) 1 2 4 3 2) 1 2 3 4 3) 2 1 4 1 4) 4 3 2 1 45) The equivalent weight of H3PO4 in the following reaction is 2 H3 PO4 + Ca(OH)2 1)

MW 1

Ca (H2 PO4 )2 + 2H2 O

 2)

MW 2

3)

MW 3

4) All

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Note : Equivalent weight of a substance depends on the reaction.

46) 300 mL of 0.1M BaCl2 is added to 100 mL of 0.1M Na3 PO4 . The amount of Ba3 (PO4 )2 precipitate formed is 1) 10-3 moles 2) 2 x 10-3 moles 3) 5 x 10-3 moles 4) 5 x 10-2 moles Hint : 3 BaCl2 + 2 Na3PO 4  Ba3 (PO 4 )2 + 6 NaCl 3 moles

2 moles

1 mole

47) Choose the incorrect statement. 1) One equivalent weight of an acid is completely neutralised by one equivalent weight of a base 2) One mole of an acid is always completely neutralised by one mole of a base. 3) One equivalent weight of an oxidising agent oxidises one equivalent weight of a reducing agent 4) All are incorrect. 48) The volume of 0.1N KOH required to neutralise completely a 20mL of 0.2N H2SO4 is 1) 10mL 2) 40mL 3) 10mL 4) 20mL Hint : Use the formula N V =N bV b a a

49) 25 mL of 0.2 M Na2CO3 is completely neutralised by 12.5 mL of HCl solution. What is the molarity of HCl solution? 1) 0.8 M 2) 0.4 M 3) 0.2M 4) 0.1M 50) 100mL of 0.2 M KMnO4 oxidises 50mL of oxalic acid solution in acidic medium completely. The molarity of oxalic acid is 1) 1M 2) 0.2M 3) 0.4M 4) 2M -2 2+ Hint : KM n O 4 + C 2O 4 + CO 2  Mn oxidant

reductant

Formula : N V = N r Vr o o N o = M o x no. of e- gained by KMnO 4 N r = M r x no. of e- lost by C 2 O 4 2-

51) 20 mL of H2SO4 is required to completely neutralise 10 mL of 0.2M NaOH solution. The weight of H2SO4 in 500 mL of H2SO4 solution is 1) 4.9g 2) 2.45g 3) 9.8g 4) None 52) 0.18gr of a metal (Atomic weight = 54g) reacts completely with 100mL of 1M HCl. The valency of the metal is 1) 1 2) 2 3) 4 4) 3 weight  w  Hint : no. of equivalents (n )= e equivalent weight(EW)

Prepared by V. Aditya vardhan [email protected] A to m ic w e ig h t E .W 

 A .W 

v a len c y

53) 'x' g of a metal 'M' is completely reacted with 100 ml of 0.2N HCl. If the atomic weight of metal is ‘100x’ g, then the formula of its sulfate salt is 1) M2 SO4

2) MSO4

3) M2 (SO4 )3

4) M(SO4 )2

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54) 20g of a bivalent metal liberates 11.2 litres of H2 gas at STP by reacting with excess of acid solution . The metal is 1) Mg 2)Al 3) Ca 4) Ba 55) 10mL of 0.2M KMnO4 solution is required to oxidise 20mL of 0.3M KI solution completely. The change in oxidation state of ‘Mn’ in this reaction is. 1) Mn7+  Mn2+ 2) Mn7+ Mn3+  3) Mn4+  Mn7+ 4) Mn7+ Mn4+  H.W : Write the balanced equation for the reaction KMnO4  6 KI  4 H 2 O  2 MnO2  3 I 2  8 KOH

56) 18g of glucose is dissolved in 1800 g of water. The mole fraction of glucose is 1) 0.1 2) 0.01 3) 0.001 4) 1 Formula : X

glucose

=

n glucose n glucose+ n water

57) The mole fraction of urea in its aqueous solution is 0.23. The mole fraction of water is 1) 0.46 2) 0.77 3) 0.23 4) 1.8 58) Which method of expression of concentration has no units and independant of temperature? 1) Normality 2) Molality 3) Mole fraction 4) All 59) The mass percentage of ethyl alcohol in water is 23%

w

. The mole fraction of alcohol is

w

1) 0.208

2) 0.104

3) 1.04

4) 3.26

Hint : solution contains 23 g ethyl alcohol + 77g water -1

60) 10 mL of a liquid (density = 4.9 g. mL ) is dissolved in 0.18Kg of water. If the mole fraction of the liquid is 0.01, then its molecular weight will be 1) 98 g 2) 49 g 3) 4.9 g 4) 490 g Hint : For very dilute solutions

nsolute X  solute nsolvent

61) The molality of an aqueous solution of urea is 3 m. What will be its mole fraction? 1) 0.3 2) 0.18 3) 0.054 4) 0.003 Hint : For dilute solutions X solute 

molality x MWsolvent 1000

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VAPOUR PRESSURE

2) 3)

4) 5)

The vapour pressure of a liquid does not depend on 1) Nature of liquid 2) Temperature 3) Surface area 4) None The rate of vapourisation of a liquid depends on 1) Temperature 2) Nature of liquid 3) Surface area 4) All The vapour pressure of a liquid at its boiling point is 1) equal to atmospheric pressure 2) Lower than atmospheric pressure 3) Greater than atmospheric pressure 4) None The vapour pressure of pure water at 373 K is. 1) 1 atm 2) 76 atm 3) 100 atm 4) 273 atm The boiling points of liquids A,B & C are 350K , 490K and 630K respectively. The correct diagram of graphs representing their vapour pressures against different temperatures.

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1)

1)

2)

C

B

vapour pressure

vapour pressure

A

T emper ature

3)

B

C

T emper ature

4)

B

A

C

B

C

vapour pressure

vapour pressure

A

Temper ature

T emper ature

Which of the following represents graph of logP against 1/T according to Clausius-clapeyron equation (where P = vapour pressure and T = temperature) 2)

log P

log P

1)

1/ T

3)

1/T

4)

log P

log P

6)

A

1/ T

1/ T

Prepared by V. Aditya vardhan [email protected] Hint : Clausius - clapeyron’s equation is ln P  

 H va p

 C

RT

H.W : What is the slope of the curve in the given graph ? Ans :

H vap 2 .3 0 3 R

Which characterizes the weak inter molecular forces of attraction in a liquid ? 1) High boiling point 2) High vapour pressure 3) High critical temperature 4) High heat of vapourisation 8) At equIlibrium, the kinetic energies of vapour and liquid are 1) Equal 2) Not equal 3) Equal only at boiling point 4) All 9) According to Raoult’s law, the partial vapour pressure of a liquid in a solution is 1) inversly proportional to its mole fraction 2) directly proportional to its mole fraction 3) equal to its mole fraction 4) greater than the vapour pressure of pure liquid. 10) According to Raoult’s law, the vapour pressure (P) of an ideal solution containing two miscible liquids i.e., A and B, is

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7)

1) P  PAo X A  PBo X B

2) P  PAo X A  PBo X B

3) P  PAo X A  PBo X B 4) All 11) A solution of two liquids behaves ideally when 12)

13)

14)

15)

1) Hmixture  0 2) Smixture  0 3) Vmixture  0 4) Hmixture  0 An ideal solution is 1) Benzene + toluene 2) n-Hexane + n-Heptane 3) Chloroben zene + Bromobenzene 4) All Which, among the following solutions, shows positive deviation from Raoult’s law 1) Ethanol + Water 2) Ether + Acetone 3) Carbontetrachloride + acetone 4) All The partial vapour pressure of liquid ‘A’ in a solution showing positive deviation from Raoult’s law is given by 1) PA = P0A XA 2) PA > P0A XA 3) PA < P0AXA 4) PA = XA The vapour pressure (P) of a solution showing negative deviation from Raoult’s law is given by (solution contains two liquids i.e, A and B) 1) P = P0A XA+ P0BXB

2) P = P0A XA - P0B XB

3) P < P0AXA+ P0B XB 4) P > P0AXA+ P0BXB 16) The solution which shows negative deviation from Raoult’s law is 1) CH3COOH + C5H5N 2) H2O + HNO3 3) CH3COCH3 + CHCl3 4) All 17) For a solution showing positive deviation from Raoult’s law 1)  H mixing  0 3) Both 1 & 2

2)  Vmixing  0 4) None

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18) The graph of vapour pressure of solution showing positive deviation from Raoult’s law is 1)

P

total

P

X

A XB

P

3)

0

0

1

0 1

total

X

A XB

4)

X

A XB

P

vapour pressure

P

0 A

1 0

P

0 A

0 B

1

P

0 B

P

P

total

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1

vapour pressure

2)

0 B

0

P

0 A

vapour pressure

vapour pressure

P

1 0

total

P

0 A

0 B

0

X

1

XB

A

1 0

H.W: Which graph shows negative deviation from Raoult’s law? Answer: 3

19) The forces of attraction in the following solution are greater than those of in their component pure liquids 1) Benzene + Toluene 2) Ethanol + Water 3) Acetone + Chloroform 4) Acetone + Ether 20) 50 mL of ethanol is mixed with 50 mL of water. The volume of the solution formed will be 1) = 100 mL 2) > 100 mL 3) < 100 mL 4) None Hint : Vmix  0

i.e., shows positive deviation from Raoult's law

21) The vapour pressure of water at 300K in a closed container is 0.4 atm. If the volumeof the container is doubled its vapour pressure at 300K will be 1) 0.8 atm 2) 0.2 atm 3) 0.4 atm 4) 0.6 atm 22) The vapour pressure of deliquescent substance is 1) Equal to the atmospheric pressure 2) Equal to the pressure of water vapour in air 3) Greater than the pressure of water vapour in air 4) Less than the pressure of water vapour in air Note : Deliquescent substance absorbs water from air and becomes liquid (by dissolving in water).

1)

2)

COLLIGATIVE PROPERTIES The property which depends only on the number of particles and not on the nature of particles is 1) Relative lowering of vapour pressure 2) Elevation in boiling point 3) Depression in freezing point 4) All Choose the incorrect statement 1) The vapour pressure of a solution containing a non volatile solute is inversely proportional to the mole fraction of solute. 2) The lowering of vapour pressure of a solution containing a non volatile solute is proportional to the mole fraction of solute. 3) The vapour pressure of a solution containing a non volatile solute is proportional to the mole fraction of volatile solvent. 4) None.

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3)

4)

5)

The vapour pressure of a dilute solution (P) containing a non volatile solute is ( Let P01 = vapour pressure of pure solvent, X1 & X2 = molefractions of solvent and solute respectively) 1) P = P01 X1 2) P = P01 (1-X2) 3) both 1 & 2 4) None The relative lowering of vapour pressure of a dilute solution containing a non volatile solute is 1) equal to mole fraction of solute 2) equal to mole fraction of solvent 3) equal to mole fraction of solution 4) greater than mole fraction of solvent 6 g of Urea is dissolved in 90g of water. The relative lowering of vapour pressure is equal to 1) 0.01 2) 0.06 3) 1.1 4) 0.02 Hint: Urea (NH 2 CONH 2 ) is a non volatile and non electrolytic substance.

Formula : Relative lowering of vapour presure 

P X2 P0

where P = lowering of vapour pressure 0

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P = vapour pressure of pure solvent X 2 = mole fraction of solute

6)

Calculate the amount of glucose present in 90g of water when the relative lowering of vapour pressure is 0.02. 1) 9g 2) 12g 3) 18g 4) 24g Hint: The molecular formula of Glucose is C 6 H 12 O 6

7)

The lowering of vapour pressure of a solution containing 15g of non electrolyte in 90g of water at 373K is 1.013 X 103N.m-2. The molecular weight of the solute is 1) 300g.mole-1 2) 90g.mole-1 3) 150g.mole-1 4) 135g.mole-1 Hint: 1 atm = 1.013 X 10 5 N.m -2

8)

1g of a non volatile and non electrolytic solute is dissolved in 78g of benzene at 780C. The molecular weight of solute is 100g.mole-1. What will be the vapour pressure of solution at 780C. 1) 75.24 mm Hg 2) 780 mm Hg 3) 684 mm Hg 4) 752.4 mm Hg Hint: The boiling point of benzene is 78 0 C

9)

10g of a non volatile and non electrolytic solute is dissolved in 1000g of benzene. The vapour pressure is lowered by 0.5%. What is the molecular weight of solute? 1) 312g.mole-1 2) 156g.mole-1 3) 78g.mole-1 4) 100g.mole-1 Hint :

P p0



0.5 100

10) The vapour pressure of an a aqueous solution of sucrose is 0.99 atm at 373 K. The molality (m) of the solution is 1) 0.99m 2) 0.55m 3) 5.5m 4) 37.3m P

Hint: p0 =X solute 

molality x MWsolvent 1000

11) The lowering of vapour pressure of 1% (w/w) aqueous solution of a non volatile and non electrolytic solute is 1.52 mm of Hg at 373K. The molecular weight of the solute is 1) 91g 2) 180g 3) 30g 4) 909g 12) The vapour pressure of a solvent is decreased by 10mm of Hg by adding a non volatile solute. The mole fraction of solute is 0.14. What would be the molefraction of solvent if decrease in vapour pressure is 20 mm of Hg 1) 0.28 2) 0.14 3) 0.72 4) 0.86 13) A 6 % glucose solution and 2 % solution of 'x' show same relative lowering of vapour pressure. Assuming 'x' to be a non-electrolyte, the molecular weight of 'x' is 1) 30g.mole-1 2) 60g.mole-1 3) 90g.mole-1 4) 180g.mole-1 Hint : Molefractions and hence the number of moles of glucose and ‘x’ are equal.

14) Correct pair of aqueous solutions with equal vapour pressures. 1) 0.1M urea, 0.2M sucrose 2) 0.1M NaCl, 0.1M glucose 3) 0.1M Na2SO4, 0.1M KCl 4) 0.2M urea, 0.1M NaCl

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15) The vapour pressure of aqueous solution of K3[Fe(CN)6] is same as that of 0.4M aqueous solution of urea. Then the concentration of K3[Fe(CN)6] is 1) 0.4M 2) 0.1M 3) 0.2M 4) 0.8M 3

 water Hint: K 3  Fe  CN 6   3 K ( aq )   Fe  CN 6  ( aq )

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16) The lowering of vapour pressure of solution containing 'X' is twice of solution containing 'Y', then (both 'X' & 'Y' are non electrolytes) 1) The MW of 'X' is twice of 'Y' 2) The MW of 'Y' is twice of 'X' 3) The mole fraction of 'X' is twice of 'Y' 4) The mole fraction of 'Y' is twice of 'X' 17) An aqueous solution of Ca(HCO3)2 is heated strongly and filtered. The vapour pressure of clear filtrate is 1) Greater than that of pure water 2) less than that of pure water 3) Equal to that of pure water 4) Can not say 

2+



Hint : Ca  aq   2 HCO3  aq   CaCO3  s    CO2  g   H 2 O( l )

18) The vapour pressure of an aqueous solution containing sucrose at 1000C is 1) = 76 cm Hg 2) > 76 cm Hg 3) < 76 cm Hg 4) = 100 cm Hg 19) Which of the following aqueous solutions has the minimum boiling point ? 1) 0.1M NaCl 2) 0.2M urea 3) 0.1M sucrose 4) 0.05M CaCl2 20) The vapour pressure of an aqueous solution of 0.2m glucose is 1atm at 100.360C. What is the molal boiling point elevation constant of the solvent ? 1) 1.8 2) 1.8 K.mole-1 3) 1.8 K.Kg.mole-1 4) 1.8 Kg-1.mole-1 Hint : Elevation in boiling point = Tb  K b . m

where K b  Molal boiling point elevation constant (or) ebbulioscopic constant m

 Molality

Note : The unit of K b is Kelvin.Kg.mole

-1

21) 0.2g of a non volatile and non electrolytic solute is dissolved in 20g of a liquid to elevate the boiling point of the liquid by 0.180C. The ebbulioscopic constant of the liquid is 4.212 K.Kg. mole-1 . What is the molecular weight of solute ? 1) 23.4g.mole-1 2) 234 g.mole-1 3) 117 g.mole-1 4) 202g. mole-1 w2 1000 ΔTb  K b . m  K b . . MW2 w 1

Hint :

Where

w 2 = wt. of solute

MW2 = Molecular wt. of solute W1 = Weight of solvent

22) The vapour pressure of a liquid becomes equal to 1 atm at -730C. The molar heat of vaporisation and molecular weight of liquid are 40 J.mole-1 and 180g.mole-1 respectively. What is it’s ebbulioscopic constant ? 1) 1.496 K.Kg.mole-1 2) 149.6 K.Kg.mole-1 3) 1.8 K.Kg.mole-1 4) 40 K.Kg.mole-1 2

 M 0

R Tb

1

Hint : K b  ΔH vap 1000 where R  Gas constant 0 Tb  Boiling point of pure liquid (solvent) ΔH vap  Molar heat of vapourisation M1  Molecular weight of liquid (solvent)

Prepared by V. Aditya vardhan [email protected]

23) 0.6 g of urea is dissolved in 1 mole of propyl alcohol. The molal boiling point elevation constant of propyl alcohol is 0.18 K.Kg.mole-1. Calculate the elevation in boiling point. 1) 0.03 K 2) 1.66 K 3) 3.3 K 4) 0.33 K 24) The aqueous solution showing maximum depression in freezing point is 1) 0. 1M CaCl2 2) 0.3M urea 3) 0.1M Na3PO4 4) All 25) 2g of non electrolytic solute is dissolved in 36g of water. The depression in freezing point observed is 1.08 K. The cryoscopic constant of water is 1.8 K.Kg. mole-1. What is the molar mass of solute. 1) 9.25g.mole-1 2) 18g.mole-1 3) 92.5g.mole-1 4) 180g.mole-1 Hint : Depression in freezing point = ΔTf  K f . m  K f .

Where

w2

MW2

.

1000 w1

Kf = molal freezing point depression constant (or) cryoscopic constant w 2 = wt. of solute

ad V. A ich DI em TYA ad VA i@ R gm DHA ail N .co m

MW2 = Molecular wt. of solute W1 = Weight of solvent

26) The latent heat of fusion of water is 3.35 X 105J.Kg-1. Calculate the depression in freezing point of 1m aqueous solution of glucose. 1) 0.00185 K 2) 1.85 K 3) 0.0185 K 4) None 2

 M 0

R Tf

  0

R Tf

1

2

Hint : K f   ΔHfusion 1000 lf .1000

  0

R Tf  T f 

2

lf .1000

 Hfusion = lf .M1 

.m

where R  Gas constant 0

Tf  Freezing point of pure liquid (solvent) ΔH fusion  Molar heat of fusion l f = latent heat of fusion

M1  Molecular weight of liquid (solvent) m = Molality

27) The freezing point of acetamide in glacial acetic acid is 298K. At this temperature, the crystals of 1) acetamide appear first 2) acetic acid appear first 3) both appear together 4) None Hint : Always crystals of pure solvent seperate out first at the freezing point of solution.

28) How many grams of glucose should be dissolved in 100g of water in order to produce a solution with a 1050C difference between the freezing point and the boiling point. (Kf = 1.86K.Kg.mol-1 and Kb = 0.51K.Kg.mole-1). 1) 18.2g 2) 37.9g 3) 180g 4) 72g T f

Hint :

T f0

Tf



 



 

0



 ΔTf +ΔTb + Tb0 -Tf0 =105 C



0

or ΔTf +ΔTb + 1000C-00C =105 C

 ΔTf +ΔTb  =50 C But ΔTf = K f m and ΔTb = K b m 0

 Kf m + Kbm = 5 C

Tb Tb0

Tb

Prepared by V. Aditya vardhan [email protected]

ad V. A ich DI em TYA ad VA i@ R gm DHA ail N .co m

29) Ethylene glycol is used as an anti freeze in radiators to 1) increase the boiling point of water 2) decrease the freezing point of water 3) decrease the boiling point of water 4) freeze the water above 00C. 30) The depression in freezing point of an aqueous solution is 0.670C. What is its relative lowering of vapour pressure?(Kf= 1.8 K.Kg.mole-1) 1) 0.0067 2) 0.042 3) 0.21 4) 0.42 31) Choose the correct statement 1) Spontaneous flow of solvent from dilute solution to concentrated solution through a semi permeable membrance is called osmosis. 2) According to van’t Hoff, the solute behaves as a gas and the osmotic pressure of solution is equal to the pressure exerted by solute if it were a gas at the same temperature and occupying the same volume as that of solution. 3) Reverse osmosis occurs, when a pressure greater than the osmotic pressure of solution is applied over it 4) All 32) The osmotic pressure of 0.2 molar solution of urea at 270C is 1) 4.92 atm 2) 1 atm 3) 2.7 atm 4) 8.2 atm Hint : Osmotic pressure ( ) = CRT where C =

n

= molar concentration

V

-1

1

R = Gas constant (0.0821 L.atm.mole .K ) T = Absolute temperature

33) 20g of a non electrolytic solute is present in 500 mL of solution at 270C. The osmotic pressure of this solution is 1.8 atm. What is the molecular weight of the solute (in g.mole-1) ? 1) 426. 2 2) 547. 3 3) 671.7 4) 323.2 34) The elevation in boiling point of an aqueous solution of glucose is 0.60C. What is its osmotic pressure at 300K ? (Kb = 1.2 K.Kg. mole-1 & density of solution = 1.09 g. mL-1) 1) 623.6 N.m-2 2) 1247.1 N.m-2 3) 600 N.m-2 4) None 35) The aqueous solution which is isotonic with 0.1M NaCl is 1) 0.1M Glucose 2) 0.1M CH3COOH 3) 0.1M KCl 4) All 36) Insulin is dissolved in a suitable solvent and the osmotic pressures (  in atm.) of solutions of various concentrations (C) in g/cc are measured at 270C. The slope of a plot of  against C is calculated to be 5 X 10-3. The approximate molecular weight of insulin is 1) 4.51 X 106 g

2) 4.926 X 106 g

Hint : π =

n W 1000 RT= X X RT V M.W V

 Slope =

π 1000 RT = C M.W

note : π =

3) 6.32 X 105g

4) 9.12 X 107g

   where C = W  in grams    V  in cc   

1000 RT W . is in y = mx form MW V

37) The correct order of osmotic pressure of equimolar solutions of BaCl2, NaCl and C6H12O6 is 1) NaCl > C6 H12 O6 > BaCl2

2) BaCl2 > NaCl > C6 H12 O 6

3) C6 H12O6 > NaCl > BaCl2 4) All are equal 38) The osmotic pressure of 0.1M NaCl is greater than that of 0.1M CH3COOH. It is due to 1) NaCl is a weak electrolyte 2) CH3COOH is acid 3) CH3COOH is a weak electrolyte 4) All 39) When mercuric iodide is added to an aqueous solution of KI, the 1) Osmotic pressure is decreased 2) Vapour pressure is increased 3) Freezing point is raised 4) All

Prepared by V. Aditya vardhan [email protected] Hint : 2KI + HgI 2  K 2 [HgI 4 ] Initially 2KI gives four ions





+ 2K and 2I

But K 2 [HgI 4 ] gives only three ions



+ 2 2K and [HgI 4 ]



40) The van’t Hoff factor (i) is given by 1)

Abnormal colligative property Normal colligative property

2)

3)

Normal molecular weight Abnormal molecular weight

4) All

Actual number of particles in solution Number of particles taken

1 Molecular weight

ad V. A ich DI em TYA ad VA i@ R gm DHA ail N .co m

Hint : Value of colligative property 

41) The relation between van’t Hoff factor (i) and degree of dissociation (  ) of a weak electrolyte ‘A’’ which dissociates by giving ‘n’ number of ions ( A  nB ) is 1)  

i 1 n 1

Hint :

2)  

i 1 n 1

A



initia l 1 mole equilibrium (1 -  )  Total no.of particles in the solution

van't Hoff factor (i) 

3)  

i n

4) None

nB

0 n = (1-  ) + n  = 1+n  -  = 1+  (n-1)





1  n 1 Total no.of particles in the solution  no.of particles taken 1

i = 1+  (n-1) i-1=  (n-1) i-1

  = n-1

42) The relation between van’t Hoff factor (i) and degree of association (  ) of ‘A’ which associates as follows nA  An . 1)  

Hint :

i 1 n 1

2)  

i 1 1 1 n

3)  



A

initial

1 mole

at equilibrium

(1- )

i 1 n 1

4)  

An 0

 n

Total no.of particles in the solution

= (1- ) +



= 1+

n

n

no.of particles in the solution  van't Hoff factor (i) = no.of particles taken

1

1+ (

-1) n

 1

i .e., i = 1+ (

1 -1) n

i-1

 = 1 n

1



i-1= (



1 -1) n

  = 1+ (

1 n

-1)

i 1 1 1 n

Prepared by V. Aditya vardhan [email protected]

43) Choose the incorrect statement. 1) Van’t Hoff factor (i) for aqueous solution of sucrose is one. 2) Van’t Hoff factor (i) for aqueous solution of acetic acid is greater than one 3) Van’t Hoff factor (i) for a solution of acetic acid in benzene is less than one 4) None Hint : Van’t Hoff factor ‘i’ = i > i < Acetic acid dimerizes

1 when there is no dissociation or association 1 when there is dissociation 1 when there is association in benzene (a solvent with low dielectric constant)

44) The degree of dissociation (  ) of a weak electrolyte AX BY is related to van’t Hoff factor (i) by the expression.

i 1 x  y 1 i 1 i 1 2)   3)   4)   x  y 1 i 1 x  y 1 x  y 1 If van’t Hoff factor of Ca(NO3)2 is 2.3, then its percentage of dissociation is 1) 22% 2) 65% 3) 60% 4) 44% The molecular weight of acetic acid dissolved in benzene is calculated as 90g.mole-1 by using osmotic pressure method. What is the percentage of association of acetic acid. 1) 0.33% 2) 0.66% 3) 90% 4) 66.6% 0 0.01m aqueous solution of K3[Fe (CN)6] freezes at -0.062 C. Its % of dissociation is (Kf =1.86K.Kg. mole-1) 1) 72% 2) 88% 3) 78% 4) 56% -1 A solution containing 0.8716 mole.L of sucrose is iso-osmotic with a 0.5M NaCl solution. What is the degree of dissociation of NaCl 1) 0.7432 2) 74.32% 3) 0.8716 4) 0.3716 The minimum value of van’t Hoff factor if a solute undergoes trimerization in a solution is 1) 0.5 2) 1.33 3) 0.33 4) 3 The observed molar mass of KCl obtained by freezing point depression method is related to the actual molar mass (M) as (the degree of dissociation is x) 1) M (1 + x) 2) M / 2x 3) M (1 + x)-1 4) M (1 - x)-1 0 At 25 C, a solution containing 0.2g of poly ethene in 100 mL of toluene shows a rise of 4.8mm at osmotic equilibrium. The molecular weight of polyethene will be (density of the solution = 0.44g.mL-1) 1) 5.29 X 103g 2) 2.98 X 105g 3) 2.39 X 103g 4) 2.39 X 105g A boy manured a rose plant with highly concentrated solution of urea. On the next day 1) the green colour of the plant will be improved. 2) the plant will grow taller. 3) the plant will bear more number of roses. 4) the plant will die.

45) 46)

47)

48)

49) 50)

51)

52)

ad V. A ich DI em TYA ad VA i@ R gm DHA ail N .co m

1)  

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