solutions hwt theory of solutions

8/12/2019 Solutions Hwt Theory of Solutions

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10VMC/Theory of Solutions Solutions to HWT-4 /Chemistry

Vidyamandir Classes

1.(C) Total mass = 120 + 1000 = 1120 g

mL1120

V1.15

Molarity =120 60

1000 2.05 M1120 1.15

2.(C) CH OH3n 5.2

Weight (g) of water = 1000 g H O21000

n18

CH OH35.2

0.08610005.2

18

3.(B) 3 3water 1kg /dm 0.3dm 0.3kg

Also urea0.01

n60

Molarity =40.01 1 5.55 10

60 (0.3)

4.(C) PV = nTRT

T41 1 n 0.082 500

nT= 1

2 6 2 2 27

C H O 2CO 3H O2

x (7/2x)

2 4 2 2 2C H 3O 2CO 2H O

(1 x) (1 x)3

7 10x 3 3x

2 3

Moles of ethane2

x3

Moles of ethane1

3

% Ethane

=2

0.673

Ethene

= 1 0.67 = 0.33

5.(D) Non-polar Benzene in water

6.(A) H = 1.0105atm.

N H N2 2P ( ) Dissolved

0.85 = 1.0105 N2( )

5N2

4 10

H2n 10

N 5 42

N24 10 n 4 10

10

7.(B) 500 g solution contain 10% of NaOH

8.(A) Compare it by formula0

100 Mm

1000d M

9.(C) Withon in temperature neither moler or mass of solvent

changes.

10.(C)29 1000

3.698 100

= 1.22

11.(C) n = 2.05 V = 1000 ml

Mass = V = 1020 gm

3CH COOHM 60 ;

3CH COOHm 60 2.05 = 123 gm

Molality =2.05

2.28 mol/ kg0.897

12.(C) For3

3 3

25 10 gmCaCO 5 ppm

5 10 ml

13.(D) 17.4 ppm means.

17.4 gm per 106parts

17.4 mg per 103parts of 1 L

3417.4 10 1

M 2 2 10174 1

14.(B) M1V

1+ M

2V

2= M

TV

T

MT 1000 = 480 1.5 + 1.2 520

MT= 1.344 M

15.(B) Moles = 103 3 21000

M 10 10100

HWT - 1Theory of Solutions


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Vidyamandir Classes

1.(D) Theory 2.(A) A B > A ~ A of B ~ B

3.(A) nHeptane

=25

0.25100

nOctane

=35

0.30114

Heptane

=0.25

0.4480.30 025

P = 105(0.448) + (1 0.448) 45

= 45 + 60 0.448 = 71.9272

4.(C) x YP 3P

5504 4

XY

P 4P 560

5 5

6X YP 400 P 600

5.(B) n - Heptane breaker H - bonds b/w C2H

5OH molecules.

6.(D) Has to be ideal solution.

7.(C) 520A+ 1000(1

A) = 760

240 = 480A

A= 1/2 or 50%

8.(C)X YP P4002 2

; X Y

P 2P350

3 3

6X YP 550 mm P 250


9.(A) 0.6 P + 0.4 200 = 290P = 350

10.(C) Apply Hesss Law

11.(D) Ben78

n 178

Tol46

n 0.594

Banzene75 1

P 501.5

12.(A) Let Weight = x.

x x PMA 4MA

x P

4MA 5

13.(A) Refer theory

14.(B) Benzene breaks H-bonds in methanol

15.(A) Refer theory

1.(A) Tf= ik

fm =

0.1/m4 1.86

100

1000

= 2.3 102K


2.(A) UseTf= ik

fm

m6 i 1.86

62

4

m = 804 . 32 g.

3.(C) = iCRT

1=

2

4.(A) Adding a non-volatile solute increase the Boiling point and

decrease the freezing point.

5.(A) Use= iCRT

i = max for Al2(SO

4)

3

7.(D) Tb= iK

bm

i = 5 for K4 [Fe(CN)

6]. Hence the boiling point will be

highest

9.(D) Calculate relative lowering and hence P.

Use BA

Pi

P

12.(B) Tf= iK

fm

0.37 = 2 Kf 0.01

x = 2 Kf 0.02

0.37 1

x 2

x = 0.74C

13.(B) Blood cells do not shrink in blood because blood is isotonic.

14.(A) Use= iCRT

15.(A) Highest boiling point will be for 0.1M FeCl3

UseTb = iKbm.


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Vidyamandir Classes


1.(B) As we know lowering of V.P. will be proportional to i.

So maximum lowerting will be observed when i is maximum.

2.(B) UseTf= iK

fm.

4.(B) P = iCRT

P = C 600 RT

2 = C/2 700 RT

P 62

2 7

24P atm

7

5.(B) = iCRT

68.41 0.082 273

342

1

= 4.48 atm

9.(D) Tf= iK

fm

There will be some lowering hence freezing point will be

below 0C

10.(D) Equimolar solutions have same boiling and same freezing

points.

11.(C) Use strength = MM0

Use= CRT

Hence osmotic pressure of solution B is greater than A

13.(A) Adding a non-voltatile solute decrease the F.P. and increase

the B.P.

14.(A) i = 4 for FeCl3.

15.(A) Maximum freezing point falls in comphor.


1.(B) Since= iCRT

i will have a larger value for KNO3. Since it is a strong

electrlyte, while CH3COOH is a weak acid.

Hence P1> P

2

2.(A) As we know BA

P

P

3.(C) Tf= iK

fm

i= 1 for glucose. So depression in freezing point will be

lowest.

Hence freezing point will be highest.

4.(A) x yA B xA yB1 0 0

1 x y

i = 1 + x+ y

i = 1 + | x + y 1)

i 1

x y 1

5.(B) UseTf= iK

fm

Kf= 1.86

m = 1 mole

i = 3

6.(A)

7.(A) UseTf= i K

fm

8.(D) 22(Ph OH) (Ph(OH)1 0

12

i 12

10.(B)2

2 4 4Na SO 2Na SO

1 0 01 2

11.(A) i1C

1= i

2C

2

(1 + 2) 0.004 = 1 0.01= 0.75

13.(D) Intermolecular forces after mixity are weaker than before

mixing.

14.(D) nA An1 0

1n

i 1

n

1i 1 1

n


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solutions hwt theory of solutions

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