Solutions Slot – 3 (Chemistry) Page # 19
EXERCISE – I SINGLE CORRECT (OBJECTIVE QUESTIONS)
7. AlCl3 = no. of particle = 4
1 1 CaCl2 = no. of particle = 3
1. y A x A
yA xA (vapour pressure of AlCl3) < (vapour pres-
sure of CaCl2 solution)
y A + yB x + xB TB.P. (AlCl3) > TB.P. (CaCl2)
< A T1 > T2
yA xA
y B xB y x 8. 100% dissociation
A A Tf = (0.0054) = i Kf m
yA xA y B xB = i × 1.86 × 0.001
=i=3
= 3 particles [MA6] A2
2. P PA PB PA0 A PB0 B
= 40 + 120 = 160 9. P = 0.95P0
PA = yA°P = yA × 160 Xsolvent = 0.95 , Xsolute = 0.005
40 = yA × 160
yA = ¼ P0 P n (Wsolute / M solute )
0
P N n Wsolute Wsolvent
P 0 P n solute n M solute M solvent
3.
P n solvent N
M solute M M solvent 0.3M
P0 P n solute
(Wsolute / M )
P 0
nsolute n solvent 0.05
Wsolute Wsolvent
M
0.3M
1 3 Wsolvent
4. xA : xB 5.7
4 4 Wsolute
1 3 10. Pure A : XB = 0
PS 100 80 85
4 4 = PT = PA° = 120
1 Pure B : XB = 1
100 = PT = PB° = 120 - 75 = 45
PA 4 25 x1
yA A
PS 85 85 11. (A) i = 1 + 0.90 (5 -1) = 1 + 3.6 = 4.6
(B) i = 1 + 0.90 (3 -1) = 2.8
60 (C) i = 1 + 0.9 (3 -1) = 2.8
yB x1B (D) i = 2.8
85 Ans : (A) Higher the value of i means Boiling
25 0 point will be higher.
Pdistilate = 100 80
85 85
= 85.88 mm Hg 0
12. P P
X solute
5. less no of particle of solute means maximum P0
vapour pressure. P 0 P X solute
6. Glucose does not dissociate 10 0.2
i=1 20 X solute
X solute 0.4 X solvent 0.6
Page # 20 Solutions Slot – 3 (Chemistry)
13. M = 1 d = 1.2 Mol mass = 180 17. PT = 200XA + 100XB
1000 M XA = XB = ½ (given)
m=
1000d M 180 PT = 150
If all the liquid become vapour then PT can be
1000 1000 calculated by
1200 180 1020
1 1 1 1
1000 yA 0 0
Tb K b 0.98 K b PT PB0 PA PB
1020
PT 133.3
actual mole of solute(Experiment)
14. i = theoritical mole of solute then pressure at which half of liquid converted
into vapour must be in the range of
MT 133.3 < P < 150
i
Mexp.
2 3
18. P = 100 × + 300 ×
i = 1 + (n - 1) = 1 + (3 - 1) 5 5
i 1 2 i 1 = 40 + 180 = 220
M
T 1
Mexp 19. For ideal solution
MT Mexp Hmix 0
Mexp 111 Gmix 0
Smix 0
15. i = 1 + (5 - 1)
= 1 + 0.6(4) = 3.4
20. Only solvent molecules can go through SPM,
Tb = 3.4 × 0.52 × 1 not the solute particles hence no blue colour
= 1.768 formation
Tb1 = 373 + 1.768
= 374.76 P0 P 1000
22. m
P Msolvent
1 1 1 1
16. P P0 y A P0 P0 17.25 17.20 1000
B A B m 0.1615
17.20 18
1/Pº
If m = M = 0.1615
B
XCl3
X 3Cl
1/P 1/Pº
A
Mtotal = S + 3S = 4S = 0.1615
S = 0.040375
yA=0 yA=1
= 4.037 × 10–2
Solutions Slot – 3 (Chemistry) Page # 21
23. Initially A = 3 mole ; B = 2 mole
1.38 10 3
3 0 2 0 26. i = 1 + (n 1)
600 = P P ...... 1
5 A 5 B
2.74 = 1 + (3 1)
finally A= 4.5 mole ; B = 2 mole and c = 0.5
mole 0.87
Degree of dissociation = 87%
4.5 0 2 0
630 = P P
7 A 7 B
27. 1.04 = 0.52 × m m = 2
PA0 = 940
P0 750 1000
0
2
P 90
B 750 18
P0 = 777 = Patm
24. B.PTolulene > B.P.Benzene P0 = Patm (because at T water boils)
V.P.Tolene < V.P.Benzene So at T : P0 = Patm
Benzene is more volatile (By graph) because when water boils V.P. become equals
xBenzene < yBenzene to atmospheric pressure
(A) If xB = 0.5 yt = 0.2, yB = 0.8
(B) xt = 0.3, xB = 0.7 yB = 0.6 not pos- 28. Freezing is exothermic because to freeze wa-
sible ter Heat must be removed.
(C) xB = 0.3 and yt = 0.4 1> False
yB = 0.6 correct 2> True.
(D) xB = 0.7. yB > 0.7 (should be)
Given yt < 0.3 yB > 0.7 correct 29. Addition of non-volatile will decrease the
freesing point and in case of addition of vola-
tile substance then it is not necessary Hence
25. 0.20 = 1.8 × M × i
Statement 1 True
0.20 20 10 Hx
i H x
Statement 2 False
0.1 1.8 18 9 0.1 0.1 0.1 0.1
10
i= and i = 1 + (2-1) = 1 +
9
1
9
1
0.1
2
0.1 81
K
1 1
1
9
