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Exercise 5

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24 views2 pages

Exercise 5

Uploaded by

Loveleen Gupta
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Page # 32 Solutions Slot – 3 (Chemistry)

Exercise - V JEE-Problems

1. 2.74 = 1 + (3  1) 6. Higher the b.p. = higher the Kb

  0.87 7. 1   2

i × 0.004 = 0.01

2. (C) I, IV i = 2.5

i = 1 + (3  1)
3. i = 1 + 0.23 (2 -1) n = 2 for CH3COOH 2.5 = 1+ 2
i = 1 + 0.23 = 1.23
   0.75
1.23  3  10  103 3 %   75%
Tf   1000  1.36
60  (500  0.997)
 0.2294
8. Tb  i K bm In acetone

4. A = 10 ; B = 12 mole initially  1.22 


 
mA  Am Let n mole remaining of A 0.17 = i1 × 1.7 ×  M  × 1000
100
n 10 - n 
solution has A = n mole M
 i1 =
B = 12 mole 122

i1 = 1
 n   12  300(n  20)
Ps=  300     500   In benzene
n  12   12  n  (n  12)

Now 0.525 mole solute added.  1.22 


 
0.13 = i2 × 2.6 ×  M  × 1000
Here solvent are (A + B)
100
Ps1 = xsolvent Ps
M 1
1
i2 = = i2 =
P = 400 (given)
s
2  122 2

ntotal = n + 12 + 0.525 In acetone Mexp = 122


In benzene Mexp = 2 × 122 = 244
n  12
xsolvent = due to dimerisation
n  12.525

 300  (20  n)   12  n  9. Tb = i Kb m


 400     n  12.525 
 12  n  
13.44
 n = 9.9 = 3 × 0.52 ×
134.4

2.303  10  = 0.156°  0.16° C


K= log   = 1.0052 × 10-4
100  9.9 

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com
Solutions Slot – 3 (Chemistry) Page # 33

72.5  nsolute 
14. Tb  K b water  n   1000
10. 7 = i × 14 ×
94  solvent  18 
i = 0.65
now water is solvent & ethanol is solute.
1 
0.65 = 1 +    1  0.1 
2 = 0.52 ×   1000  3.21
0.9  18 

  0.7
Tb = 100 + 3.21 = 103.21° C
%   70% = 376.21 K.

2 phenol  phenoldimer
15. Pair = KH xair
 5 = 105 × xair
1 
2
0.3 xair = 5 × 10-5
0.35
dissolved in water
35 %
nair (in water )
 5  10 5
nair(in water )  nwater
20 1000
11. 2 = i × 1.72 × 
172 50 nwater = 10
nair(in water) = 5 × 10-4
i=½
nN2 (in water) = 5 × 10-4 × 0.8

12. Tf  K f m = 4 × 10-4

 n 
 K f  solute  1000
 w solvent 

 nsolute 
 Kf   1000 
 nsolvent  w solvent 

 0.1 
 2  1000  
 0.9  46 

= 4.831
Tf = 155.7 - 4.831 = 150.87 K
In solution ethanol is solvent & water is sol-
ute.

13. Ps = 0.9 × 40
= 36
(water is solute considered as non-volatile ac-
cording to question)

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com

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