Solutions Hwt Theory of Solutions

Vidyamandir Classes

Theory of Solutions 1.(C)

Total mass = 120 + 1000 = 1120 g

HWT - 1 6.(A)

 1120  VmL     1.15 

2.(C)

PN 2   H ( N 2 ) Dissolved 0.8  5 = 1.0  105 ( N 2 )

120 60 Molarity =  1000  2.05M 1120 1.15



n H2  10

Weight (g) of water = 1000 g

3.(B)

4.(C)

CH3OH 

 n H2 O



1000  18

5.2  0.086 1000 5.2  18

water  1kg / dm3 

0.3dm3  0.3kg

Also

 0.01  n urea     60 



0.01 1   5.55  104 Molarity = 60 (0.3)

 N2 10

 4  105  n N 2  4  104

7.(B)

500 g solution contain 10% of NaOH

8.(A)

Compare it by formula m 

9.(C)

With  on  in temperature neither moler or mass of solvent

100 M 1000d  M 0

changes. 10.(C)

11.(C)

PV = nTRT

 29  1000 3.6      98   100      



 = 1.22

n = 2.05 V = 1000 ml

41  1  n T  0.082  500

Mass = × V = 1020 gm

nT = 1

M CH3COOH  60 ; m CH3COOH  60  2.05 = 123 gm

7 C2 H 6  O 2   2CO 2  3H 2 O 2



x

(1 – x) (1 –

For CaCO3 

13.(D)

17.4 ppm means. 

2 Moles of ethane x  3

Ethane =

Ethene = 1 – 0.67 = 0.33

M  2

14.(B)

 5 ppm

17.4 gm per 106 parts 17.4  103 1   2  104 174 1

M1V1 + M2V2 = MTVT MT × 1000 = 480 × 1.5 + 1.2 × 520

Non-polar Benzene in water

MT = 1.344 M 15.(B)

VMC/Theory of Solutions

5  103 ml

17.4 mg per 103 parts of 1 L

1 Moles of ethane  3

2  0.67 3

2.05  2.28 mol / kg 0.897

25  103 gm

12.(C)

x)3

7 10 x  3  3x  2 3

%

Molality =

(7/2x)

C2 H 4  3O2   2CO2  2H 2 O

5.(D)

 N2  4  105

n CH3 OH  5.2



H = 1.0  105 atm.

10

Moles = 10–3

M  103 

1000  102 100

Solutions to HWT-4 /Chemistry

Vidyamandir Classes

Theory of Solutions 1.(D)

Theory

3.(A)

25  0.25 nHeptane = 100

nOctane =

A  B > A ~ A of B ~ B

2.(A)

9.(A)

10.(C)

Apply Hess’s Law

11.(D)

n Ben 

0.25  0.448 0.30  025

n Tol 

 P = 105(0.448) + (1 – 0.448) 45 = 45 + 60 × 0.448 = 71.92  72 4.(C)

 Px 3PY   550 4 4

5.(B)

n - Heptane breaker H - bonds b/w C2H5OH molecules.

6.(D)

Has to be ideal solution.

7.(C)

520 A + 1000(1 – A) = 760 240 = 480 A A = 1/2 or 50% P P 400  X  Y 2 2

PBanzene 

x x  P MA 4MA

x P  4MA 5

13.(A)

Refer theory

14.(B)

Benzene breaks H-bonds in methanol

15.(A)

Refer theory

1.(A)

Tf = ikfm = 4  1.86 

P 2P 350  X  Y 3 3

;

75  1  50 1.5

Let Weight = x. 

 PX  400 PY6  600

78 1 78

46  0.5 94



12.(A)

 PX 4   PY  560  5 5

8.(C)

0.6 × P + 0.4 × 200 = 290 P = 350

35  0.30 114

Heptane =

HWT - 2

0.1/ m 100 1000

= – 2.3 × 10–2 K

 PX  550 mm PY6  250

Theory of Solutions 2.(A)

Use Tf = ikfm

6  i  1.86 

 3.(C)

4.(A)

HWT - 3 9.(D)

m 62 4

m = 804 . 32 g.

Use

12.(B)

 PA

 i B

Tf = iKfm 0.37 = 2 × Kf × 0.01

1 = 2

x = 2 × Kf × 0.02

Adding a non-volatile solute increase the Boiling point and

0.37 1  x 2 x = 0.74°C

Use  = iCRT i = max for Al2(SO4)3

7.(D)

P

 = iCRT

decrease the freezing point. 5.(A)

Calculate relative lowering and hence P.

Tb = iKbm i = 5 for K4 [Fe(CN)6]. Hence the boiling point will be

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 FeCl 3 Use Tb = iKbm.

highest

VMC/Theory of Solutions

11

Solutions to HWT-4 /Chemistry

Vidyamandir Classes

Theory of Solutions 1.(B)

As we know lowering of V.P. will be proportional to i.

HWT - 4 9.(D)

Tf = iKfm

So maximum lowerting will be observed when i is maximum.

There will be some lowering hence freezing point will be

2.(B)

Use Tf = iKfm.

below 0°C

4.(B)

P = iCRT

10.(D)

points.

P = C × 600 RT 2 = C/2 × 700 × RT

11.(C)

Use strength = MM0 Use  = CRT

P 6  2 2 7 P

Equimolar solutions have same boiling and same freezing

Hence osmotic pressure of solution B is greater than A

24 atm 7

13.(A)

Adding a non-voltatile solute decrease the F.P. and increase the B.P.

5.(B)

 = iCRT

68.4  1  0.082  273 = 4.48 atm 342 1

14.(A)

i = 4 for FeCl3.

15.(A)

Maximum freezing point falls in comphor.

Theory of Solutions 1.(B)

HWT - 5

Since  = iCRT

6.(A)

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

7.(A)

Use Tf = i Kfm

8.(D)

 (Ph(OH) 2 2(Ph  OH)  1 0  1  2

electrlyte, while CH3COOH is a weak acid. Hence P 1 > P 2 2.(A)

As we know

3.(C)

Tf = iKfm

P  PA

 B

i 1

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

10.(B)

Hence freezing point will be highest. 4.(A)

 xA  yB A x B y  1 0 0 1  x y

11.(A)

 5.(B)

 

i 1 x  y 1

Na 2SO 4   2Na   SO 42 1 0 0 1  2  i1C1 = i2C2 (1 + 2) × 0.004 = 1 × 0.01  = 0.75

i = 1 –  + x + y i = 1 + | x + y – 1)

 2

13.(D)

Intermolecular forces after mixity are weaker than before mixing.

14.(D)

Use Tf = iKfm Kf = 1.86

 An nA  1 0  1  n

i  1  

 n

1  i  1    1 n  

m = 1 mole i=3

VMC/Theory of Solutions

12

Solutions to HWT-4 /Chemistry