. pm = 10 M ass of the solution - Karnataka Examination...

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n 1. Molarity = 2 L n v 2. Molality = 2 n ( ) 1 K g w 3 P 6 10 w × 3. Ppm = 2 10 w M ass of the solution × 4. Mole fraction of solute = 2 1 2 n n n +

Transcript of . pm = 10 M ass of the solution - Karnataka Examination...

n1. Molarity = 2

L

nv

2. Molality = 2n

( )1 K gw

3 P 61 0w ×3. Ppm = 2 1 0wM a s s o f t h e s o l u t i o n

×

4. Mole fraction of solute =2

1 2

nn n+

5. Mole fraction of solvent =1

1 2

nn n+

6. Mass percentage of solute = 2 1 0 0wM a s s o f s o l u t i o n

×

7 ) . . . . . . .A H AP K x= H ’ l f l bilit

•P =0 0

A A B Bx p x p+ ‐ Ideal binary solution of liquidsRaoult’s law.

7 ) . . . . . . .A H AP K x Henry’s law of gas solubility

02

01 2

p p np n n−

=+

•Raoult’s law of the solution ofa nonvolatile solute.

1 In 91% Ethanol by mass the mole1. In 91% Ethanol by mass, the molefraction of Ethanol is :

a) 0.8 b) 0.7 c) 0.9 d) 0.6

Mole fraction of Ethanol =

9146 2

0 8= =+46

91 946 18

0.82 5

ANS : ( a)+46 18 2.5

2) If ‘x’ molal solution of a compound in benzene has mole fraction of solute benzene has mole fraction of solute equal to 0.2, the value of ‘ x ‘ is :

a) 14 b) 3.2 c) 1.4 d) 2.0

→ =1000

1 12.5 .78

x moles Kg of Benzene moles

M.F. of Solute = ⇒ =0.2 . . 0.8M F of Benzene

→0.8 0.2moles of Benzene moles of solute

12.5 moles ×→

0.2 12.53.2

ANS (b)

→ 3.20.8

ANS : (b)

3) The volumes of 4 M HCl and 10M HClrequired to prepare 1 litre of 6 M HCl is :required to prepare 1 litre of 6 M HCl is :

a)0 75L of 4 M HCl + 0 25L of 10 HCla)0.75L of 4 M HCl + 0.25L of 10 HCl

b)0.67L of 4 M HCl +0.33L of 10M HCl

c) 0.25L of 4 M HCl +0.75L of 10 M HCl

d) 0.80L of 4 M HCl + 0.20 L of 10 M HCl

( )+ − = ×4 10 1 6 1x x

∴2

x∴ =3

xANS : (b)ANS : (b)

4) Which of the following aqueous Solution should have the highest boiling g gpoint ?

a)1 M NaOH

b) 1 M Na2SO4

c) 1 M NH4NO3

d) 1 M KNO3d) 1 M KNO3

ANS : bANS : b

5. The values of Vant Hoff’s factor for5. The values of Vant Hoff s factor forKCl, NaCl and K2SO4 are respectively

) 2 2 b) 2 2 3 a) 2,2,1 b) 2,2,3

) 1 1 2 d) 1 2 1c) 1, 1, 2 d) 1, 2,1

ANS : (b)

6) 4 L of 0.02M solution of NaCl was dilutedby adding 1L of water. The molarity of thesolution is :

a) 0.004 b) 0.008

c) 0.012 d) 0.016

=1 1 2 2M V M V× = ×

1 1 2 2

20 .0 2 4 5M 2

∴ =2 0.016MANS : ( d )

7) The Molarity of HCl if the density of 7) The Molarity of HCl if the density of the Solution is 1.825g/cc is :

)25 b) 50 ) 30 d) 40a)25 b) 50 c) 30 d) 40

1 825

− =1.825

36.53 50

1 10Molarity =

−× 31 10

ANS : ( b )

8) 12g of urea is dissolved in 1L of the Solution and 68.4g of Sucrose was dissolved in 1L

f h l i Th l i l i f of the solution. The relative lowering of vapourpressure of urea solution is :

a) greater than Sucrose solutionb) d bl th t f S l tib) double that of Sucrose solutionc) less than Sucrose solutiond) l t th t f S l tid) equal to that of Sucrose solution

Molarity of urea = =12

60 0.21

My1

68.4Molarity of Sucrose = =342 0.2

1M

ANS : (d )

9) Ethylene glycol is used as antifreeze in a cold climate Mass antifreeze in a cold climate. Mass of ethylene glycol to be dissolved in 4 Kg of water to prevent it from 4 Kg of water to prevent it from freezing at -5.60 will be

1⎡ ⎤

a) 500g b) 624g

11.86fK K Kg mol−⎡ ⎤=⎣ ⎦

a) 500g b) 624g

c) 744 g d) 800gc) 744 g d) 800g

∆ = ×f fT K m

= ×5.6 1.86 m

= ⇒3. 3 /m moles Kg

12 / 4or moles Kg/o moles g

× =12 62 74Kg=2 62CH OH M2

2CH OH

ANS : ( c )

10) If Sodium sulphate is completely ionized in solution, the value of

fT∆

when 0.01 mol of Sod sulphate is dissolved in 1 kg of water is

( )1.86fK =

a) 0.0744 b) 0.0186

c) 0.0372 d) 0.0558

∆ =f fT iK mf f

= × ×3 1.86 0.01 = 0.0558

ANS : ( d )

11)The elevation in the B.pt of a solutionof 13 44g of CuCl in 1kg of water isof 13.44g of CuCl2 in 1kg of water is( Kb=0.52K kg mol-1 and Mol mass of CuCl =134 4)CuCl2=134.4)

a)0.05 b) 0.1 c) 0.16 d) 0.21) ) ) )

∆ =b bT i K m∆ b bT i K m1344

= × × =13.44

3 0.52 0.161344 1×134.4 1

ANS : ( c )( )

12)The molarity of the resulting S l ti bt i d b i i 4L f 0 2 M Solution obtained by mixing 4L of 0.2 M NaOH with 2L of 0.5M NaoH is

a) 0.9 M b) 0.3M

c) 1.8M d) 0.18 M

( ) ( )4 02 2 05MV MV ( ) ( )× + ×+= =

+1 1 2 2 4 0.2 2 0.5

0.36

MV MVM

V V+1 2 6V V

ANS : (b)( )

13)Conc HNO3 is 70% by mass. How many 13)Conc HNO3 is 70% by mass. How many gms of conc HNO3 should be used to prepare 250mL of 2M HNO3 ?p epa e 50 o O3 ?

)45 b) 90 ) 70 d) 54a)45g b) 90g c) 70g d) 54g

Molarity = = × ××

22 2

2L

L

Wor W M olarity M V

M V2 L

= × × =2 63 0 25 31 5g= × × =2 63 0.25 31.5g100

315 45M f HNO = × =3 31.5 4570

Mass of HNO g

ANS : (a)

14) 5 ml of acetone is mixed with 100ml of H O The vapour pressure of waterof H2O. The vapour pressure of waterabove the solution is

a)equal to the V.P. of pure water) f b)equal to the V.P. of solution

c) less than the V.P. of pure waterd) h h f Od)More than the V.P. of pure H2O

ANS CANS : C

15) At a given temp, the vapour pressures of two liquids A and B are 120 p qand 180 mm of Hg respectively. The V.P. of the ideal solution containing 2 moles of gA and 3 moles of B will bea)156mm b) 145mma)156mm b) 145mm

c) 150mm d) 108mmc) 150mm d) 108mm

= +0 0A A B BP X P X PA A B B

⎛ ⎞ ⎛ ⎞= × + × =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 31 2 0 1 8 0 1 5 6

5 5m m

⎝ ⎠ ⎝ ⎠5 5

ANS ( )ANS : ( a )

16) Negative deviation from Raoult’s law isobserved in which of the following binaryliquid mixtures ?

a) Ethanol & acetonea) Ethanol & acetoneb) Benzene & Toluenec) Chloro Ethane & Bromo Ethanec) Chloro Ethane & Bromo Ethaned) Acetone and Chloroform

ANS : dANS : d

17)18g of glucose is added to 178.2 g of 17)18g of glucose is added to 178.2 g of water. The vapour pressure of water forthis solution at 1000C isthis solution at 100 C is

a)76 Torr b) 752 4 Torra)76 Torr b) 752.4 Torr

c) 759 Torr d) 7 6 Torrc) 759 Torr d) 7.6 Torr

18g of Glucose = =1 8

0 . 11 8 0

m o l e

178.2g of water = =1 7 8 . 2

9 . 91 8

m o l e s

− −∴ = ⇒ =

0

0

0 .1 7600 .01

10 760P P P

P

∴ = − =7 6 0 7 . 6 7 5 2 . 4P

ANS : b

18.The osmotic pressure of a solution8. e os ot c p essu e o a so ut oat 00C is 4 atmos.The osmotic pressureat 546 K is

a) 4atm b) 2atmc) 8atm d) 1atm

Tπ 1 1TT

ππ

=1 1

Tπ 2 2

ANS : c

19) An aqueous solution of KI is 1 molal.Which of the following will cause gthe vapour pressure of the solution toincrease?

a)Addition of waterb)Addition of NaClb)Addition of NaClc) Addition of Na2SO4d)Addition of Kl solutiond)Addition of Kl solution

ANS : ( a )

20) Th ti f 20) The average osmotic pressure of human blood is 7.8 bar at 370C. What is th C n f N Cl th t ld b i j t d the Concn of NaCl that could be injected into the blood ?

)0 15M b) 0 30Ma)0.15M b) 0.30M

) 0 60M d) 0 45Mc) 0.60M d) 0.45M

78π 7.8 0.15192 0.083 310

iCRT CiRTππ= ∴ = = =

× ×

ANS : ( a )

21) A solution of urea boils at 100.180C at1 atmos. If Kf and Kb for water are 1.86 f band 0.52 K kg Mol-1 respectively, theabove solution will freeze at

a)-6.540C b) -0.6540C

c) 6.540C d) 0.6540C

186 T∆f fK T

K T∆

=∆

1.86 0.65052 018

fT∆= =

b bK T∆ 0.52 0.18

ANS : b

22 Solutions A B C and D are espectively 22.Solutions A, B, C and D are espectively 0.1M Glucose, 0.05M NaCl, 0.05M BaCl2 and 0 1M AlCl3 Which one of the BaCl2 and 0.1M AlCl3. Which one of the following pairs is isotonic ?

a)A & B b) B & C

c) A & D d) A & C

ANS : (a )

23) 0.1M NaCl and 0.05M BaCl2 are separated by S P M which of the separated by S.P.M. which of the following is true ?

a) No movement across the membraneb) Water flows into NaCl solutionb) Wate ows to aC so ut oc) Water flows into BaCl2 solu tiond) Osmotic Pressure of 0.1M NaCl is) Os ot c ess e o 0. aC s

lower than that of BaCl2

ANS : (c )

24) If two liquids A and B have Then the molefraction of A in Vapours is

0 0: 1:2A BP P =

Then the molefraction of A in Vapours is(M.F’s of A & B are in the ratio of 1:2)

a)0.33 b) 0.25

c) 0.52 d) 0.2

0 0A A B BP x p x p= +A A B BP x p x p+

( ) ( )1 1 2 2× + ×= 5

( ) ( )1 1 2 2× + ×

1PM.F. of A in vapour phase =

1 0.25

APP= =

ANS ( d )ANS : ( d )