Solutions
11.1
ANSWERS
Section A
Level-I
1. Non-ideal solution with positive deviation.
2. Relative lowering of vapour pressure.
3. Barium chloride furnishes three moles of ions in solution while sodium chloride furnishes two moles
of ions.
For BaCl2, i = 3
and for NaCl, i = 2
4. As the van’t Hoff factor increases, boiling point increases and the freezing point decreases. The
increasing boiling point order is
C2H5OH < Na2SO4 < Li3PO4 < Ba3(PO4)2
The increasing freezing point order is
Ba3(PO4)2 < Li3PO4 < Na2SO4 < C2H5OH
5. Let the amount of Na2CO3 = x g
then, the amount of NaHCO3 = (1 – x)g
But, nNa2CO3 = nNaHCO3 (given)
1– x
x
=
100
84
The reactions involved are:
\
Na 2 CO3
0.00528 mol
NaHCO3
0.00548 mol
+
2 HCl
x = 0.56 g
\
nNa2CO3 = 0.00528 and nNaHCO2 = 0.00548
® 2 NaCl + H 2 O + CO 2
2 × 0.00528 mol
+
HCl
® NaCl + H 2 O + CO2
0.00548 mol
\ number of moles of HCl required = 2 ´ 0.00528 + 0.00548 = 0.01604
But 0.01604 = Molarity ´ volume of HCl
\
VHCl = 0.1604 L = 160.4 mL
6. poHex = 105.2 kPa
poHep = 105.2 kPa
Vapour pressure of hexane, pHex = p°Hex xHex
nHex =
25.0
= 0.29
86
and
0.29
= 0.45
and
0.29 + 0.35
Vapour pressure of hexane, pHex = poHex xHex
\ xHex =
Chapter-11 OLC.p65
1
nHep =
35.0
= 0.35
100
xHep =
0.35
0.29 + 0.35
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11.2
Physical Chemistry for IIT-JEE
= 105.2 ´ 0.45 = 47.34 kPa
and pHep = p°Hep xHep
= 46.8 ´ 0.55 = 25.47 kPa
The total vapour pressure of the mixture = 47.34 + 25.74 = 73.08 k Pa
7. Let the atomic mass of A be x
and the atomic mass of B be y
then, molecular mass of AB2 is x + 2y and that of AB4 is (x + 4y)
In case of AB2
WB = 1g
WA = 20 g
DTf = 2.3 K
and kf = 5.1°C Kg mol–1
then (x + 2y) =
5.1 ´ 1.0 ´ 1000
2.3 ´ 20
x + 2y = 111
For the solute AB4 , we have
...(1)
5.1 ´ 1.0 ´ 1000
1.3 ´ 20
x + 4y = 196
Solving equation (1) and equation (2) give
y = 42.5
and
x = 26.0
Hence, Atomic mass of A = 26.0 g mol–1
and atomic mass of B = 42.5 g mol–1
(x + 4y) =
...(2)
36
= 0.2 mol
180
Molarity of glucose solution = 0.2 M
\ we know that F = M R T
8. At 300 K, nglu =
4.98
= 0.083 L bar mol –1 K –1
0.2×300
Now, if the osmotic pressure is 1.52 bar, then the concentration is
\
R=
M=
9. Wsugar = 34.2 g
1.52
F
= 0.061 M
=
RT 0.083× 300
nsugar = 0.1 mol
\ nH2O = 10 mol
(i) Molality, m =
nsugar
WH 2O
´ 1000 =
WH2O = 180
0.1
×1000
180
m = 0.555 m
Chapter-11 OLC.p65
2
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11.3
Solutions
(ii) xsugar =
nsugar
n sugar + n H2 O
=
0.1
= 9.9×10–3
0.1+10
10. First of all we calculate the molarity of the HCl used for the neutralization of Na2CO3 solution.
n Na
2 CO 3
=
10.6
= 0.1 mol
106
Molality of Na2CO3 solution =
0.1
= 1M
0.1
Number of moles of Na2CO3 neutralised by HCl = MNa2Co3 ´ VNa 2 CO3
1.0 ´ 25 ´ 10–3 = 2.5 ´ 10–2 mol
The reaction involved is
Na 2 CO3 + 2 HCl ® 2 NaCl + H 2 O + CO 2
\ number of moles of HCl = 2 ´ 2.5 ´ 10–2 = 5.0 ´ 10–2
Molarity of HCl used for the neutralization of Na2CO3
5.0 ´ 10–2
= 1.0 M
50 ´ 10 –3
Initial HCl was 10 M, we have to prepare 500 cm3 of1.0 M HCl. This is a dilution problem, therefore,
MHCl =
M1 V1 = M2 V2
10 ´ V1 = 1.0 ´ 500
V1 = 50 cm3
11. density = 1.0097 Kg dm–3
It means 1dm3 solution weigh 1009.7 g which contains 80.8 g ethanoic acid.
\
Weth = 80.8 g
and
WH2O = 1009.7 – 80.8 = 928.9 g
(i) mass percent of water =
928.9 ´ 100
= 91.99%
1009.7
and mass per cent of ethanoic acid =
(ii) neth =
80.8
= 1.35 mol
60
nH
80.8 ´ 100
= 8.01%
1009.7
2O
=
928.9
= 51.61 mol
18
n H 2O
51.61
=
Mole fraction of water, xH2O = n
51.61+1.35
H 2 O + n eth
xH2O = 0.975
and mole fraction of ethanoic acid = 1 – 0.975 = 0.025
(iii) Molality of ethanoic acid =
Chapter-11 OLC.p65
3
1.35
´ 1000 = 1.45 m
928.9
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11.4
Physical Chemistry for IIT-JEE
(iv) Partial vapour pressure of ethanoic acid, peth = p°eth xeth
peth = 100 × 0.025 = 2.5 mmHg
(v) Total vapour pressure, p = peth + pH2O
\
pH2O = 125 – 2.5 = 122.5 mmHg
(vi) Partial vapour pressure of water, pH2O = poH2O xH2O
\
Vapour pressure of pure water =
p H 2O
x H 2O
=
122.5
= 125.64 mmHg
0.975
12. First solution
\
nA = 1 mol
and
nB = 2 mol
xA = 0.33
and
xB = 0.67
Then, 300 = p °A ´ 0.33 + p°B ´ 0.67
Second solution
...(i)
xA = xB = 0.5, therefore,
350 = 0.5 p°A + p°B ´ 0.5
...(ii)
solving equation (i) and (ii) gives
p°B = 203 mmHg
p °A = 497 mmHg and
13. WB = 1.00 g
MB = 172 g mol–1
WA = 10.0 g
MA = 58 g mol–1
p° = 400 mmHg
p =?
\
nA =
10.0
1.00
= 0.172 and n B =
= 5.81×10 –3 mol
58
172
Now, the mole fraction of solute, xB =
5.81 ´ 10–3
0.172 + 5.81 ´ 10–3
and
x B = 0.033
We know that,
\
p = (1 – xB) ´ p° = (1 – 0.033) ´ 400
p = 386.8 mmHg
V = 1 dm3
14. WB = 5.0 g
p = 0.096 kPa = 9.50 ´ 10–4 atm
T = 288 K
We know that the molecular mass of protein can be calculated as
MB =
WB RT 5.0 ´ 0.0821 ´ 288
=
pV
9.50 ´ 10–4 ´ 1
M B = 1.24 ´ 105 g mol –1
15. WB = 5.0 g
p = 4.16 kPa
Chapter-11 OLC.p65
WA = 100 g
and
4
p° = 4.24 kPa
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11.5
Solutions
\ Relative lowering of vapour pressure =
We know that RLVP =
or
But
1+
nB
nA + nB
nA
1
=
= 53
n B 0.019
0.107 =
p° – p 4.24 – 4.16
=
= 0.019
p°
4.24
WB
\
MB
\
nA
= 52
nB
or
MB =
16. WB = 7.39 g
nB
nA + nB
0.019 =
\
nB =
100
= 0.107
18 ´ 52
5.0
= 46.73 g mol –1
0.107
WA = 85.0 g
DTb = 355.6 – 353.2 = 2.4 K
kb = 2.53 K kg mol–1
We know that MB =
\
MB =
k b WB 1000
D Tb WA
2.53 ´ 7.39 ´ 1000
= 91.65 g mol –1
2.4 ´ 85.0
17. Calculation of boiling point of solution:
Given :
WB = 0.300 g
\
nB =
0.300
= 5.0×10 –3 mol
60
and molality of the solution, m =
5.0 ´ 10–3 ´ 1000
= 0.5 m
10.0
Now, DTb = kb m
DTb = 0.52 ´ 0.5 = 0.160 K
\ Boiling point of solution is 373.160 K
Calculation of osmotic pressure:
Mass of the solution is 10.3 g
\ Volume of the solution is
\ Molality of the solution =
10.3
= 9.45 mL
1.09
5.0×10–3
= 0.529 M
9.45×10–3
We know that F = M R T = 0.529 ´ 0.0821 ´ 373.160 = 16.21 atm
18. DTf = 273.0 – 268.5 = 4.5 K
number of moles of KCl =
Chapter-11 OLC.p65
5
10.0
74.5
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Physical Chemistry for IIT-JEE
We know that, DTf = iK f m
D T f ´ WH2 O
or i =
k f ´ n KCl ´ 1000
=
4.5 ´ 100 ´ 74.5
= 1.80
1.86 ´ 10.0 ´ 1000
0.13 ù
é 3.5
´2+
´ 3ú = 0.122 mol
19. Number of moles of solute, nB = ê
95
ë 58.5
û
mass of water = (100 – 3.5) + (100 – 0.13) = 196.37 g
We know that DTb = kb m
0.122 ´ 1000
or D Tb = 0.323
196.37
The boiling point of solution is 100.323°C or 373.323 K.
DTb = 0.52 ´
Level-II
1. Density of C2H5OH = 0.98 g/mL and mass of C2H5OH = 0.98 ´ 100 = 98 g and number of moles of
98
C2H5OH =
= 2.13 mol suppose the mass of glucose added = x gm then, number of moles of
46
n glu
x
mol and the mole fraction of glucose is 0.02 =
glucose =
n glu + n C2 H5 OH
180
50 = 1+
or
n C2 H5OH
and n glu =
n glu
2.13
= 0.043
49
x
\ x = 7.82 gm
180
Hence, 7.82 gm of glucose should be added to 100 mL C2H5OH in order to have a mole fraction of
0.02.
2. First we calculate the molarity of 98% by mass sulphuric acid. 98% by mass means 98g H2SO4 present
in 100 gm solution.
0.043 =
But,
\
n H SO =
2
4
98
= 1.0 mol
98
100
1
mL = L
1.80
18
1.0
So, molarity of the solution is
= 18M
1/18
Similarly, we can find the molarity of a 23% by mass H2SO4 solution,
and volume of the solution is
n H SO =
2
\
Chapter-11 OLC.p65
4
23
= 0.235 mol
98
Molarity =
6
volume =
100
1
mL = L
1.20
12
0.235
= 2.82 M
1/12
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11.7
Solutions
This is a dilution problem, therefore,
Mconc. Vconc. = Mdil. Vdil.
Volume of 98% by mass H2SO4 required,
2.82 ´ 500
= 78.33 mL
18
Calculation of mass of water added
Mass of the 23% solution prepared = 500 × 1.20 = 600 gm
Mass of the 98% H2SO4 added = 78.33 × 1.80 = 141 gm
\
mass of water added = 600 – 141 = 459 gm
3. 1.25 M Na2SO4 means 1.25 mol Na2SO4 in 1L solution.
V=
n Na
2SO4
= 1.25 mol
and WNa SO = 1.25×142 = 177.5 gm
2
4
mass of solution = dsol ´ Vsol = 1.25 ´ 1000 = 1250 gm
\
and
WH O = 1250 –177.5 = 1077.5 gm
2
n H 2O =
1072.5
= 59.58
18
(i) Molality of solution w.r.t. Na+ ions =
n Na +
WH2 O
´ 1000
(2 ´ 1.25)
´ 1000
1072.5
= 2.33 m
=
(ii) Mole fraction of Na2SO4 = x Na 2SO4 =
n Na 2SO4
n Na 2SO4 + n H2 O
1.25
1.25 + 59.58
= 0.021
=
(iii) Mass percent of Na2SO4 =
mass of Na 2SO4
×100
mass of solution
177.5
´ 100
1250
= 14.2 %
4. 23% w/v means 23 gm H2SO4 is present in 100 mL solution or 100 mL solution contains 23 gm H2SO4.
and 1L (or 1000 mL) solution contains 230 g H2SO4.
As density of solution is 1.20 g/mL, therefore, the mass of solution is 1200 g.
\ 1200 g solution contains 230 g H2SO4
=
230
= 2.35 mol
98
and mass of water = 1200 – 230 = 970 g
Then
Chapter-11 OLC.p65
n H 2SO 4 =
7
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11.8
Physical Chemistry for IIT-JEE
Therefore, the molality of the solution, m =
2.35 ´ 1000
= 2.42 m
970
5. Glauber’s salt is Na2SO4.10H2O
Given:
Wsolute or WB = 8.06 ´ 10–2 kg = 80.6 g
Vsolution = 1 dm3 = 10–3 m3
dsolution = 1077.2 Kg m–3
\
mass of solution = d ´ V = 1077.2 ´ 10–3 Kg = 1077.2 g
80.6
= 0.568 mol
142
(i) \ Molarity of Na2SO4 solution is 0.568 M
(ii) Mass of water = 1077.2 ´ 80.6 = 996.6 g
Number of moles of Na2SO4 =
Therefore, the molality of the solution, m =
0.568
´ 1000
996.6
m = 0.570 m
(iii) number of moles of Glauber salt = 0.568 mol
nH2O =
996.6
= 55.37 mol
18
Therefore, the mole fraction of Na2SO4 =
6. WDCM = 60.0 g
nDCM =
0.568
= 0.010
0.568 + 55.37
WDBM = 30.0 g
60
= 0.71
85
WDBM =
30.0
= 0.17
174
Now, the mole fraction of dichloromethane, x DCM =
=
n DCM
n DCM + n DBM
0.71
= 0.807
0.71 + 0.17
\ x DBM = 1 – 0.807 = 0.193
vapour pressure of dichloromethane, pDCM = p°DCM ´ xDCM
= 0.175 ´ 0.807 = 0.141 atm
and pDBM = p°DBM ´ xDBM = 0.015 ´ 0.193 = 2.895 ´ 10–3 atm
(i) Total vapour pressure of the solution = 0.141 + 2.895 ´ 10–3 = 0.144 atm
(ii) Mole fraction of dichloromethane in the vapour phase is
yDCM =
pDCM 0.141
=
= 0.98
p total
0.144
7. Given : mass of benzene = 56.8 × 0.889 = 50.50 gm
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11.9
Solutions
p° = 100 mmHg
p = 98.88 mmHg
and
\
Relative lowering of vapour pressure, RLVP =
=
p° – p
p°
100 – 98.88
= 0.0112
100
For dilute solution,
We know that, Molality, m =
(RLVP) ´ 1000
MA
0.0112 ´ 1000
= 0.144 m
78
We know that, DTf = k f m
m=
D Tf 0.73
=
= 5.07 kg mol –1
m
1.44
8. First of all we calculate the number of moles of the solute using
\
kf =
F=
n B RT
V
F V (527 / 760) ´ 0.1
=
= 2.83 ´ 10–3
RT
0.0821 ´ 298
Now, suppose the mass of sugar (lactose) is x g, then the mass of the drug will be (1 – x )g. But
\
nB =
1– x
x
+
= 2.83 ´ 10–3
342 369
Solving for x gives x = 0.58
\
mass percent of sugar =
0.58
´ 100 = 58%
1.0
9. K 2CrO 4 ® 2 K + + CrO42–
\
i=3
M = 0.1 M
T = 298 K
We know that
F = i M R T = 3 ´ 0.1 ´ 0.0821 ´ 298 = 7.34 atm
10. Given :
Wprotein = 0.225 g
V = 10.0 mL = 10.0 ´ 10–3 L
p = 0.257 atm
T = 298 K
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Physical Chemistry for IIT-JEE
(a) We know that, molar mass of solute is given by
Mproten = Wprotein RT
FV
0.225 ´ 0.0821 ´ 298
= 2142 g mol –1
0.257 ´ 10.0 ´ 10 –3
(b) Actually, protein dissociates to produce 20 Na+ ions and one p20–.
Therefore, i factor = 21
We know that,
normal molecular mass
i=
abnormal molecular mass
\ actual (or normal) molecular mass of protein
= 21 ´ 2142 = 44982 g mol–1.
=
Chapter-11 OLC.p65
10
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