1.
ADDITIONAL PROBLEMS ON SOLUTIONS
SOLUTIONS
Some more Additional
Problems on Solutions (Solved)
1. What is the molarity of an aqueous solution of ethyl alcohol (CH3 CH2 OH, molar
mass = 46.0 g/mol) which contains 50.0 g of alcohol in 500 mL of the solution?
Solution.
50 1000
=100g
Mass of ethyl alcohol/ litre =
500
Mass in g / litre
100
=
=2.18
No. of moles of alcohol/ litre = =
Molar mass in g /mol 46
Thus, the solution contains 2.18 moles of ethyl alcohol per litre of the solution. So, the
molarity of the solution is 2.18 mol / L.
2. Calculate the normality· and molarity of H2S04 in a solution containing 9.8 g of H2SO4
per dm3 of the solution.
Solution:
Therefore,
Hence,
Mass of H2S04 per litre = 9.8 g
Molar mass of H2SO4 = 2 x (1) + 32 + (4 x 16) = 2 + 32 + 64 = 98 g mol-1
98
Number of moles of H2S04 per litre of solution =
=0.1
98
Molarity of H2S04 in solution= 0.1 moll 1-1
As there are two equivalents per mole of H2 SO4, therefore
Normality of H2S04'in solution= 0.1 x 2 = 0.2.equiv L-1
3. 2.82 g of glucose (molar mass180 g mol-1) are dissolved in 30 g of water Calculate.
(i) molality of the solution (ii) molefractions of, (a) glucose (b) water.
Solution.
Mass of glucose,
w8= 2:82 g
Molar mass of glucose,
Mg = 180 g mot-1
Mass of water,
ww = 30 g
BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL
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2.
ADDITIONAL PROBLEMS ON SOLUTIONS
Molar mass of water, Mg = 18 g mol-1
(i) From the definition, molality is the number of moles per 1000 g of the solvent. Then,
2.28
Number of moles of glucose m 30 g water =
mol = 0.0.157 mol
180
0.0157
1000 =. 0.522' mol kg-1
So,
Molality of glucose =
30
(ii) From above, the number of moles of glucose (ng) and. water (nw) in the solution are
given by,
Mass of glucose
282
n
g
=
Molar mass of glucose
Mass of water
180
= mol =· 0 0157 mol
30
=
mol = 1.667mol
nw
Molar mass of water 18
ng
Therefore, Molefraction of glucose =
ng + n w
0.0157
=
0.0157+1.667
ng + n w
0.01
=
0.99
1.667
nw
Molefraction of water =
=
=
1.667+0.0157
4. What is the molality of the aqueous solution of methyl alcohol (CH3 OH. molar mass
32.0 g/mol) which contains 64 g of methyl alcohol in 200g of water?
Solution:
Mass of methyl alcohol in 200 g of water = 64 g
64 1000
Then,
=320g 1
Mass of methyl alcohol in
200
320g
So,
No. of moles of alcohol per 1000 g of water
32g/ mol
=
10 mol
Thus, I kg of the solution contains 10 mole of methyl1 alcohol. So, the molality of the
solution is 10 mol /kg.
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3.
ADDITIONAL PROBLEMS ON SOLUTIONS
5. An agueous of solution of a. dibasic acid (molar mass=118 g/mol) containing 17.7 g
of the acid per litre of solution has a density, I. 0077 g/mL. Express the concentration of
the solution in as many as you can
Solution.
Mass of solute,
Molar mass" of the solute,
w
M
= 17.7 g/L
= 118 g mol-1
Mass of solute (g/L)
So
'No ·of moles of solute dissolved =
w
=
-1
Molar mass (g mpl ) M
mol/L
17.7
Then Mass of the solvent, (water)
So,
Volume of the solution
Density of solution
Mass of solvent (water)
Then, Maas of the solvent, (water)
N o. of moles of solvent
Total number of moles in solution
=
Mol /L = 0.15 mol /L
118
= 1000 mL ·
= 1.0077 g / mL
= Volume x Density = 1000 mL x 1.0077g/mL
= 1007.7 g
= Mass of solution - Mass of solute
= (1007.7- 17.7 g) = 990 g
Moles of solute
990
=
=
=55.0
Molar mass in g / mol
18
= Moles of solute + Moles of solvent
= 0.15 +55 = 55.15
Since, the acid is dibasic, hence its
Eq. mass = Molar mass/ 2 =59 g equiv -1
From the above results the concentration of solution can be calculated in various
units as follows:
(i) Molarity: Already calculated above : Molarity = 0.15 mol/L
(ii) Molality: From above,
1000g
1
Mass of solute (acid) per 1000 g of solvent =17.7
=0.151mol / kg
990 g 118
Mass in g per 1000 g solvent
No. of moles of the solute/ 1000 g of solvent =
1000
1
= 0.151 mol / kg
=17.7
990
So,
Molar mass (g/ mol)
118
Molality = 0.151 mol/ kg
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4.
ADDITIONAL PROBLEMS ON SOLUTIONS
n
(iii) Mole- fraction
(X) =
n
acid
0.15
n
acid+ water
=
0.15
=
0.15+55
=0.027
55.15
6. How much sodium chloride be dissolved to make l litre of 0.1 F solution?
Solution:. Sodium chloride is an ionic .compound and is represented by the formula
NaCl.
its gram-formula mass is, 23 g + 35.5 g
= 58.5, g
Therefore,
So,
1 litre of 1 F NaCl solution, one requires
= 58.5 g, of NaCl
1 litre of 0.1 F NaCI solution, one. requires
= 58.5 x O.1g = 5.85 g
Thus, 5.85 g of NaCl should be dissolved to make up 1 litre of solution.
7. How many grams of KCI would be required to prepare 1 litre of 0.1 M
solution? Atomic masses are; K= 39 u, Cl-= 35.5 u
Solution: Potassium chloride (KCl) is an ionic compound with a molar mass of
39 + 35.5 = 74.5 g mol-1 Therefore,
For 1 litre of 1 M KCl solution, one .requires
74.5 g of KCl
74.5 0.1
For 1 litre of 0.1 M KCl solution, one requires
= 7.45g
1
Thus, one requires 7.45 g of KCl to prepare 1 litre of its 0.1 M solution
8. 2.46 g of sodium hydroxide (molar mass = 40 g/mol) are dissolved in water and the
solution is made to .100 mL in a volumetric flask. Calculate the molarity of the
solution.
Solution:
Mass of sodium hydroxide
= 2.4'6 g
Molar mass of sodium hydroxide ·= 40 g moll-1
Volume of the · solution
= 100 mL
Molarity of the solution
=?
Molarity of a solution is given by the number of moles of solute present in 1000 mL of
the solution, i.e.,
No. of moles of solute
Molarity =
Volume of the solution in mL
x 1000 ml L-1
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5.
ADDITIONAL PROBLEMS ON SOLUTIONS
(Mass of the solute I Molar mass of the solute)
=
x 1000 mol L -1
Volume of the solution in mL
2.46/ 40) 1000
=
2.46
molL-1=
100
-1
10molL- = 0.615molL
40
So, the molality of the solution is 0615 mol L-1
9. Calculate the molality of a 1 M solution of sodium nitrate. The density of the solution
is, the solution is 1.25 g cm-3
(NaN03) = (23 + 14 + 48) g/mol = 8S
Solution: Molar mass of sodium nitrate,
g/mol
Mass of 1dm3 (or 1 litre) of the solution = Volume x Density Therefore,
= 1000 cm3 x1.25 g/cm3 = 1250 g
Therefore,
Mass of water containing 8.5 g .of NaN03 = (1250- 85) g = 1165 g = 1.165 kg
So,
1mol
-1
Molality (m) of the solution. =
=0.86 mol kg
1.165kg
10. 0.75.g of sodium bicarbonate (NaHC03) are dissolved in250 ml of a solution.
Calculate its, (i) normality, (ii) molarity
Solution.
Mass of sodium bicarbonate dissolved = 0.75g
Volume of solution
= 250 mL
Molar mass of sodium bicarbonate . = (23+ 1 + 12+~8) g mol-1 = 0.615 L
1
g mot'
Since, one molecule of NaHCO3 contains only one cationic charge (on Na+), hence
Molar mass
Equivalent mass, (E)=
84
=
1
g equiv
-1
1
As per definition,
(Mass of NaHCO 3 / Molar mass of NaHC0 3 )
Molarity (M) =
Volume of solution in mL
-1
0.75g/84 mol
0.75/84mol
1000mL
=
250/1000L -1
250 mL
x 1000
-1
= 0.0357molL
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6.
ADDITIONAL PROBLEMS ON SOLUTIONS
(Mass of NaHCO 3 / Equiv. mass of NaHC0 3 )
Normality (N) =
1000 ML / L
Volume of solution in mL
11. Calculate the molefraction of water in a mixture of 12 g water .108 g acetic acid and
92 g ethyl alcohol.
Solution:
Following the procedure of the previous problem, one can write
Mass of water 12
= =0.67
n(H2O)
=
Molar mass
18
Mass of ethanol
92
=
= 2.00
n(C2H5OH) =
Molar mass
46
Mass of acetic acid
108
n(CH3COOH) =
=
Molar mass
=1.80
60
So,
Total number of moles in the solution, ntotat = 0.67 + 2.00 + 1.80 = 4.47
Therefore,
X water =
0.67
=
n total
=0.15
4.47
n total
nC2H5 OH
X ethanol =
Xacetic acid =
nH2 O
2.00
=
=0.45
4.47
nCH3 cooH
n total
1.80
=
=0.40
4.47
12. What is the molality of ammonia in a solution containing 0.85 g NH3 in a 100 cm3of a
liquid of density 0.85 g cm-3?
Mass of ammonia
w2 = 0.85 g
Molar mass of ammonia,
M2=17 g mol-1
Mass of liquid (solvent),
Solution:
w1= 100 cm3 0.85g cm-3=0.85 g
BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL
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7.
ADDITIONAL PROBLEMS ON SOLUTIONS
So,
Molality of ammonia
n
NH
w /M
-1
3
= 2 2 mol kg
=
Mass of solvent in kg 85/1000
or,
Molality of ammonia
=
0.85/17
mol kg-1 = 0.59 mol kg-1
85/1000
13. An aqueous solution of a dibasic acid (C2H2O4.2H2O) contains 1.26 g of the solute
per litre of the solution Calculate the normality end molarity of the solution
Solution: Mass of the acid W=1.26 g/L
=
Molar mass M of dibasic acid
2 12 + 2 1 + 4 16 + 2 2 +
16
-1
= 24 + 2 + 64 + 36 = 126 gmol
So,
Equivalent mass of the acid =
Molar mass
2
=
126
2
= 63 g equiv
-1
According to the definition,
Molarity
=
and, Normality
Mass of solute in g per L
Molar mass
=
=
1.26 g / L
126 g/mol
Mass of solute in g per L
Equivalent mass
=
-1
= 0.01mol L
1.26 g / L
63 g equiv -1
-1
= 0.01mol L
14. What volume of 95% sulphuric acid (density=1.85 g/cm3) and what mass of water
must to taken to prepare 100 cm3 of 15% solution of sulphuric acid (density)
=1.10 g /cm3)?
Solution:
Volume of the solution
Density of the solution
Therefore
= 100 cm2
=1.10 g /cm3
Mass of 100 cm3 of solution = 100 x 1.10 g =110 g
The given solutions is 15%. This means that 100 g of solution contains 15 g of H2 SO4
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8.
ADDITIONAL PROBLEMS ON SOLUTIONS
Then, mass H2SO4 in 110 g (=100 cm3) of solution =
15g
100g
110g = 16.5g
and
Mass of water in 110 g (=100cm3 ) of solution = =(110-16.5) g =93.5 g
So,
to obtain 100 cm3 of 15 % solution acid, we require
Mass of water
= 93.5 g
Mass of H2SO4 (100% pure)
=16.5 g
Since, the given sulphuric acid is 95% pure, hence
Mass of H2SO4 (95%) required =
Density of 95% H2SO4
100 16.5g
95
= 17.37g
=1.85 cm-3
15. Sea water contains 6 x 10-3 g of dissolved oxygen in one litre Express the
concentration of oxygen in sea water per million (ppm) units
Density of seawater
=1.03 g /mL.
Solution: Mass of 1 littre of sea water = 1000 mL x 1.03 g /mL =1.030 g
Mass of dissolved oxygen in sea water per litre = 6 x 10-3 g
6 10-3
So,
Concentration of dissolved oxygen =
106 ppm = 5.8 ppm
1030 g
16. Calculate the molality of a salt (molar mass = 138 g mol-1) solution obtained by
dissolving 2.5 g in one litre of the solution. Density of solution is 0.85 g cm-3
Solution: Mass of the salt dissolved
=2.5 g
Volume of the solution
=1000 cm3
Density of the solution
=0.85 g cm3
So, Mass of 1000 cm3 of solution
Therefore
Thus
=Volume x Density= (1000 x 0.85) g= 850 g
Mass of solvent = (850-2.5) = 847.5 g
847.5 g of solvent contains 2.5 g of solute
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9.
ADDITIONAL PROBLEMS ON SOLUTIONS
Or,
2.5
847.5 g of solvent contains
1 g solvent contains =
1.38
2.5
1
1.38
847.5
1000 g of solvent contains =
mol of solute
mol
1000
2.5
8475
138
= 0.0214 mol
Therefore, molality of the solutions is 0.0214 mol kg-1
17. Calculate the mass of the solute present in the following solutions:
(a) l L of N/10 Na2CO3 solution
(b) 2 L of N /10 HCI solution
(c) 100 mL of 0.5 M H2SO4
(d) 250 mL of N/10 oxalic acid
Solution.
(a) Preparation of 1 L of N/ 10 solutions:
Molar mass of Na2 CO3= (2 23) + 12 + (3 16) = 106g mol -1
We know, that the gram equivalent mass of Na2Co3 is half of the molar mass Hence,
Equivalent mass of Na2 CO3= =
106
= 53 g equiv -1
2
Thus 1 L of 1 N Na2CO3 solution should contain 53 g of Na2 CO3
Then, 1 L of N/ 10 Na2CO3 solution would contain
53
10
g of Na2 Co3 = 5.3 g Na2CO3= 5.3
g Na2CO3
(b) Preparation of 2 L of N/ 10 HCI solution:
Equivalent mass of HCI = (1+35.5) = 36.5 g equiv-1
So,
1 L of 1 N HCl solution should contain 36.5 g of HCI. Then,
BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL
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10.
ADDITIONAL PROBLEMS ON SOLUTIONS
2. L of N /10 HCl solution would contain =
36.5
2
10
73
g of HCl =
10
g HCl =7.3 g HCl
(c) Preparation of 100 mL of M H2 SO4
Molar mass of H2 SO4 contains 98 g H2 SO4
Or
1000 mL of 1 M H2SO4 contain 98 g H2SO4
Then, 100 mL of 0.5 M H2SO4 would contain =
98
1000
100
0.5
1
g of H2SO4
= 4.9 g of H2SO4
(d)Preparation of 250 mL of N/10 oxalic acids:
Oxalic acid crystals correspond to the formula, (COOH)2 2 H2O
Molar mass of oxalic acid crystals = (2 12+4 16 +2 1) +2 (2 +1 +16)
=(24 +64 + 2 +36) =126 g mol-1
Basicity of oxalic acid =2
So,
Equivalent mass of oxalic acid =
Then,
126
= 63 g equiv -1
2
1 L of 1 N solution of oxalic acid contains 63 g oxalic acid
1000 mL of 1 N
250 mL of
N
10
“
“
“
“
63 g oxalic acid
”
“
“
“
=
=
63
1000
63
1000
250
1
10
g oxalic acid
= 1.575 g oxalic acid
18. Find the molarity and molality of a 15 % solution of H2SO4 (density of H2SO4
solution =1.10 g/ cm3 Molar mass of H2SO4 98 g mol-1)
Solution: 15% Solution of H2SO4 means that 100 g of solution should contain
15 g of H2SO4 So.
Mass of H2SO4 dissolved
=15 g
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11.
ADDITIONAL PROBLEMS ON SOLUTIONS
Mass of solution
Density of the solution
=1.10 g cm-3
Molar mass of the H2SO4
So,
=100 g
=98 g mol -1
Mass of water containing 15 g of H2SO4 = (100 – 15 )g =85 g
No. of moles of H2SO in 15 g of H2SO =
and, Volume of 100 g of solution =
15 g
98 g/ mol
mass
Density
=
15
98
mol = 0.153 mol
100g
=
1.1g cm
-3
= 90.9 cm
3 90.9 dm3
1000
(i) Calculation of molarity
No. of moles of H2SO4
Molarity of H2SO4 solution =
Volume of solution in dm
0.153 mol
=
3
(90.9 / 1000) bm
3
=1.68 mol dm
-3
So, thee molarity of H2SO4 solution is 1.68
(ii) Calculation of molality
Molarity of H2SO4 solution =
No. of moles of H2SO4
Molality = =
Mass of water in g
0.153
85
1000mol kg
-1
1000
= 1.8 mol kg
-1
19. Urea forms an ideal solution in water Determine the vapour pressure of an aqueous
solution containing 10 per cent by mass urea at 40oC (Vapour pressure of water of 40oC
=55.3 Hg)
Solution:
Concentration of urea
=10%
Let
Mass of the solution
=100g
Then
Mass of urea
=10 g
Therefore,
Mass of water
=(100-10)g = 90g
Molar mass of urea (NH2CONH2)
=60g mol-1
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12.
ADDITIONAL PROBLEMS ON SOLUTIONS
Then,
Mole- fraction of urea,
Xurea =
Xurea =
Then, from Raoult’s law,
nurea
nurea + nwater
0.1667
5.1667
=
10/60
(10/60 + 90/18
=
1.1667
0.1667 + 5
= 0.032
po - p o
= Xurea
po
55.3 - Ps
= 0.032
55.3
So, the vapour of urea solution at 40o C is 53.5 mm Hg.
20. The vapour pressure of water at a creation temperature is 18.15 torr .and that of a
solution containing 9.47 g of sugar, at the same temperature , is 18.06 torr Calculate
the molar mass of the sugar.
PoA=18.15 torr
Solution Given Vapour pressure of water (solvent)
Vapour pressure of solution,
PA=18.06 torr
Mass of sugar (solute)
W B = 9.47 g
Molar of water (solvent)
WA=100g
Molar mass of water (solvent)
MA=18 g /mol
Molar mass of sugar(solute),
MB = ?
From Eq (3.11),
MB =
WB MA
WA
PA
9.47 18
=
o
P A -P
A
100
18.03
18.15-18.06
WB MA
Form Eq. (3.13) M
=
B
WA
o
P A
o
P A -P
A
g / mol =342 g/ mol
9.47 18
=
100
18.03
18.15 -18.06
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13.
ADDITIONAL PROBLEMS ON SOLUTIONS
21 The vapour pressure of a 5% aqueous solution of a non – volatice organic substance
at 373 K is 745 mm Hg. Calculate the molar mass of the substance.
Solution.
From the given data, for 100 g solution,
Mass of Solute,
W2 = 5 g
Mass of solvent (water)
W 1=(100-5) g= 95 g=0.095 kg
Vapour pressure of solution, Ps =745 mm Hg
Vapour pressure of pure solution.(water) at373 K, Po s=760 mm Hg
M2=?
Molar mass of solution,
M1 =18 g/ mol
Molar mass of solvent (water),
P
Form Raoult,s law,
- Ps
n2
W2 / M2
s
=
=
o
n1 +n 2 W 1/ M1+ W
/M
P s
2
2
760- 745
5/M2
=
o
o
5/M2
=
P s 760 95 /18+ 5/M
2
5.278 +5/M
2
The above equation may be rewritten as
760 5.278 +5 / M2
=
15
5/M
2
5.278 M
=
2
+1=1.0556M2 +1
5
50.67 =1.0556 M2 + 1
-1
50.67-1
or
M
=
2
1.0556
1.0556
=
47g mol
22. Calculate the boiling point of a one molar aqueous solution (densitry: 1.03 g mL-1) of
sodium chloride. Kb for water = 0.52 K kg mor-1Atomic mass. Na = 23.
CI= 35.5.
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14.
ADDITIONAL PROBLEMS ON SOLUTIONS
So,
= 1 molar = 1 mol L-1
Conc. of the solution
Density of solution
= 1.03 · g mL-1
Molar mass of NaCl
= (23 + 35.5) g/mol = 58.5 g/ mol-1
Mass of 1 litre of solution = 1000 x 1.03 =1030 g
Therfore,
Mass of water containing 1 mole of NaCI =(1030 - 58.5)g = 971.5 g
Solution.
1 1000
mol/kg
=
Thus,
Molality, of the solution, m =
Then,
So,
1.0293 mol/kg
ΔTb= I Kb m = (2 0.52 1.0293) K= 1.07 K
Boiling point of solution = (373.15 +1.07) K = 374.22 K
971.5
23. An aqueous solution containing 2.4 g of a substance per 100 g solvent shows an
elevation of boiling point of 0.21 K what is the molar mass of the substance? Kb for
water (solvent) is 0.52 k kg mol-1
Mass of solute.
w2 =2.4 g
Mass of solvent,
Solution:
w1=100g =0.1 kg
ΔTb=0.21 K
Molar mass of the solute=? (M2)
0.52 2.4
Kb w2
M2=
we know,
w1 Tb
=
0.1 0.21
g/mol = 59.4 g/ mol
24. A solution contains 3.5 g of a non-volatile solute in 125 g of water; and it boils at
373.52 K. Calculate the molar mass of the solute. (Kb for water= 0.52 K/m)
Solution·
Let,
w2 = 3:5 g
Mass· of the solute,
Mass of water,
w1 =125 g = 0.125 kg
Elevation of boiling point, Δ Tb = (373.52- 373.0) K = 0.52 K.
Molar mass of the solute
= M (?)
We know that,
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15.
ADDITIONAL PROBLEMS ON SOLUTIONS
Tb =
K b n2
Kb w
2
=
w1
/ M 0.52 kg mol
-1
3.5g
M 0.125 kg
w1
-1
0.52 3.5 K kg mol g
0.52 K=
M 0.125 kg
-1
0.52 3.5 K kg mol g
M=
3.5
-1
-1
g mol = 28 g mol
=
0.52 K 0.125 kg
0.125
25. The boiling point of water becomes 100.52 C, if 1.5 g of non-volatile solute is
dissolved in 100 mL of it. Calculate the molar mass of the solute. Kb for water = 0.6 k/m.
Solution.
Mass of solute = 1.5 g
Volume of water (solvent) = 100 mL
Taking density of water as 1 g/mL,
Mass of water (solvent) = 100 mL x 1 g/mL = 100 g = 0.1 kg·
ΔTb = (100.52 - 100) C = 0.52 C
or
Kb
w2
w1
Kb w 2
We know,
M
ΔTb =
M=
0.6 1.5
Tb w1
0.52 0.1
g/ mol = 17.3 g/ mol
26. A solution of 3.795 g sulphur in 100 g carbon disulphide (boiling point, 46.30 C)
boils at 46.66 C what is the formula of sulphur molecule in the solution? Kb for carbon
disulphide is 2.42 K kg mol-1
Solution.
Mass of sulphur (solute) = 3.795 g
Mass of carbon disulphide (solvent) = 100 g = 0.1 kg
Molar mass of sulphur = M
ΔTb = (46.66 - 46.30) C = 0.36 C
We know,
M=
Kbw2
w1 Tb
2.42 3.795
=
g/mol =255.1g/mol
0.1 0.36
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16.
ADDITIONAL PROBLEMS ON SOLUTIONS
Atomic mass of sulphur = 32 g/mol
So,
255.1
No. of sulphur atoms in one molecule =
=7.97 =8
32
So, sulphur exists as S8 in the solution.
27. A solution .of urea in water freezes at 0.400 C. What will be the boiling of the same
solution if the depression and elevation constants for water are 1.86 deg kg mol-1 and
0.5 12 deg kg mol-1 respectively?
Solution:
We have the relationships,
K b .n 2
K f .n 2
Tb
….(i)
and Tf
…(ii)
w1
w1
where, n2 is the number of moles of the solute, and w1 is the mass of solvent in kg.
Dividing Eq. (i) by Eq. (ii)
K
T
b
b
=
T
f
or
b=
T
Then,
K
Tf K b
f
0.4 0.512
=
o
o C = 0.11 C
1.86
Kf
Boiling point of the urea solution = 100o C + 0.11oC = 100. 11oC
28 .The normal freezing point of nitrobenzene, C6H5 NO2 is 248.82 K .A 0.25 molal
solution of certain in nitrobenzene causes a freezing point depression of 2 degree.
Calculate the value of Kf for nitrobenzene.
Solution:
Normal freezing point of nitrobenzene =278.82 K
Concentration of solution
=0.25 molal =0.25 mol/kg
Freezing point depression Tf = 2K
K f ?
We know that,
Tf =K f m
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17.
ADDITIONAL PROBLEMS ON SOLUTIONS
Tf
-1
2K
=
Kf
m
=8 K kg mol
0.25 mol/kg
29. Find the (i) boiling point, and (ii) freezing point of a solution containing 0.520 g
glucose (C6H12 O6) dissolved in 80.2 g of water for water. Kf =1.86 K/m, and Kb =0.52
K/m.
= 100oC
= 0.0oC
= 0.52 g
=180 mol
80.2
Mass of water =80.2 g =
kg
1000
K n
b 2
We know =
T
where, n2 is no. of moles of solute (w2/
w
b
M2) w1 is the mass of solvent in kg
1
Solution:
Normal boiling point of water
Normal freezing point of water
Mass of glucose
Molar mass of glucose (C6H12O6
=
Tb
0.52 0.52
0.52 (w 2 /M2 )
=
K = 0.019K = 0.02K
w1
180 (80.2/1000)
So,
Boiling point of solution = (373+ 00.2) K =0.067 K
K (w /M )
1.86
0.52
2 2
Similarly = f
T
= 0.067 K
180
(80.2/1000)
w
f
1
So, Freezing point of solution = (273 K – 0.067 K)
30. Ethylene glycol (HoH2C- CH2OH) is used as an antifreeze for water to be used in
car radiators in cold places .How much ethylene glycol should be added to 1 kg of water
to prevent it from freezing at -10o? Molal depression constant of water is 1.86 k kg mol-1
Solution:
Mass of water (solvent)
w1=1 kg
o
Tf =10 C
K f =1.86 K kg mol
-1
w2=?
Mass of ethylene glycol required,
K f (w 2 / 62)
We know,
T
f
=
w1
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18.
ADDITIONAL PROBLEMS ON SOLUTIONS
1.86 w 2
10
=
62 1
10 62 1
This gives
w
=
g
1.86
2
=
333.3g
31 A solution containing 4 g of a non- volatile organic solute per 100 cm3 was found to
have an osmotic pressure equal to 500 cm Hg. at 27oC. Calculate the molar mass of
the solute.
Solution:
Osmotic pressure
= (5000/ 76) atm
Temperature, Mass T = (27 +273) =300 K
of solute, Volume W= 4 g
of solution,
V=100 cm3 atm deg-1 mol-1
R = 0.0821 dm3atm deg-1 mol-1
w
pv =
We know,
RT
M
w
This gives,
M=
V
4 0.0821 300
RT
500 / 76
g/mol
=
149.7g/mol
0.1
32. 5 g of a non – volatile non- electrolyte solute is dissolved in water and the solution
was made up to 250 cm3. The solution exerted an osmotic pressure equal to 4X105
N M-2 at 298 K Find the molar mass of the solute
Solution:
Mass of solute,
Volume of solution,
Osmotic pressure
Temperature,
Molar mass of solute
w=5g
= 0.005 kg
3
V= 250 cm =250 10-6 m3 =2.5 10-4 m3
=4 105N m-2
T=298 K
M =?
R= 8.314 j k-1 mol-1
wRT
We know,
or
M
=
V
0.005 kg 8.314 JK
=
5
-2
4 10 N m
-1
m,ol
-1
2.5 10
-4
298 K
m
3
M=0.1239 kg mol-1 = 123.9 g mol-1
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19.
ADDITIONAL PROBLEMS ON SOLUTIONS
33. Osmotic pressure of a solution containing 7.0 g protein in 100 m L of solution is 20
mm Hg at 37oC Calculate the molecular mass of the protein .(R = 0.0821 litre
atmosphere deg-1 mol-1 )
Solution:
Mass of protein,
w=7.0 g
Volume of the solution,
V = 100 mL = 0.1 L
Molecular mass of the protein = M
Temperature, T= 37o = (37+273) k=310 K
20
Osmotic pressure, =20 mm Hg
760
atm
we have,
wRT
M=
7.0 0.0821 310
=
7.0 0.0821 310 760
=
=67700 g / mol
(20/760) 0.1
pV
20 0.1
34. At 298 K, 100 cm2 of a solution containing 3.02 g of an unidentified solute
exhibits an osmotic pressure of 2.55 atmosphere. What is molecular mass of solute?
(R=0.0821 Ll atm mol-1 K-1)
Solution:
Given:
T= 298 K
Volume of the solution
Mass of Solute,
Osmotic pressure,
V=100 cm3 =0.1 L
W= 3.02 g
=2.55 atm
w
The osmotic pressure is given by, V=n RT =
wRT
So.
M
=
pV
M
RT
3.02 0.0821 298
=
2.55 0.1
=
67700 g / mol
35. A 5% solution of cane sugar C12 H22 O11 is isotonic with 0.8 77 % of solute A
Calculate the relative molar mass of A. Assume density of solutions to be 1 g cm-3
Solution: Conc. of sugar solution =5%
Mass of solution
=100g
So, Mass of sugar in solution = 5 g
Relative molar of sugar = (12 12+22+11 16) =144+22+176=342
We know,
nv = nRT
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20.
ADDITIONAL PROBLEMS ON SOLUTIONS
or,
n
=
w RT
RT=
V
M
V
5
So,
and,
342
Sugar =
RT
V
A=
0.877 RT
M
V
Since, the two solutions are isotonic, hence,
0.877
M
or,
Sugar =
RT
So,
5
V
0.877 342
M=
A,
=
RT
342
V
= 60g/mol
5
36. Calculate the molecular mass of a substance 1.0 g of which on being dissolved in
100 g of solvent gave an elevation of 0.307 K in the boiling point. ( Molal elevation
constant, Kb =1.84 K/m).
K b n2
Solution:
We know, Tb =
where n2 is the no. of moles of solute; w1 is
w1
the mass of the solvent in kg.
Mass of solute
1.0 g
=
n2 =
Molar mass of the solute
M
and
w1 =100 g =0.1kg
Substituting these values, one can write,
1.84K kg mol
0.307K
=
-1
0.1Kg
1.84K kg mol
-1
1.0 g
M
1.0g
1.84 1.0
So, M =
g/ mol= 59.9 g/ mol
0.1Kg 0.307 K
0.1 0.307
37. The density of water is room temperature is 0997 g cm3 Calculate the molarity of
pure water.
Solution: The density of water is 0.997 g cm3. Thus,
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21.
ADDITIONAL PROBLEMS ON SOLUTIONS
Mass of 1 cm3 of water =0.997 g
Therefore, Mass of 1 dm3 =1000 cm3 of water =0.9997 g cm-3 1000 cm3= 997 g
Molar mass of water= 2+16=18 g mol-1
No. of moles of water per dm3 55.39
Therefore,
Therefore, the molarity of water is 55.39 mol L-1
38. A solution is 25 percent water 25 percent ethanol and 50 percent acetic acid by
mass. Calculate the mole- fraction of each component.
Solution: Let us consider 100 g of the solution .Then,
mw=25 g
Mass of water,
Mass of ethanol,
m eth =25g
Mass of acetic acid macet=50 g
Molar mass of water, ethanol and acid are, 18 g/ mol, 46 g / mol, and 60 g/
mol respectively Then,
25g
No. of moles of water,
nw=
= 1.39
18 g/mol
25g
No. of moles of ethanol,
n eth=
No. of moles of acetic acid
n acet=
Total no of moles in solution
60g/mol
=(1.39 +0.54+0.83) =2.76
46g/mol
50g
= 0.54
=0.83
1.39
So,
Mole –fraction of water,
Mole- fraction of ethanol,
Mole- fraction of acetic acid
Xw=
2.76
= 0.50
0.54
=0.20
2.76
0.83
=0.30
Xacet =
2.76
Xeth =
39. Concentrated sulphuric acid has a density of 1.9 g/ ML and is 99% H2SO4by
weight Calculate the molarity of H2SO4 in this acid.
Solution: Let us consider 1 litre of H2SO4Then
Mass of 1 litre of this acid =1000 mL 1.9 g/mL =1900 g
Since, the given acid is 99% pure, hence
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22.
ADDITIONAL PROBLEMS ON SOLUTIONS
1900
99
Actual mass of H2SO4 in 1 litre sample =
= 1881g
100
Molar mass of H2SO4=(2+32+64)g mol =98 g mol-1
-1
1881g
So, No of moles of H2SO4 in 1 litre of the sample =
98 g mol
= 19.19 mol
-1
So, the given H2 SO4 sample is 19.9 molar.
40. Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of
nitric acid Calculate the volume of the solution which contained 23 g of HNO3.
Density of the conc. HNO3 Solution is 1.41 g cm-3.
Solution: Let, Mass of conc. HNO3 sample =100g
So,
and
Mass of HNO3 in 100 g of sample = 69 g
Mass of HNO3 in 100 g of sample = 31 g
Density of conc. HNO3 sample
=1.41 g cm-3
Mass
So, Volume of 100 g of the HNO3 sample =
Dwnsity
2
100g
=
1.41g
3
Thus, 69 g of HNO3 is contained in .70.92 cm of conc. HNO3
cm-3
= 70.92cm
70.92
1g
“
“
23g
“
“
69
70.92
“
23 “
=23.6 cm3
69
Thus, 23.6 cm3 conc. HNO3 sample contained 23 g HNO3
41. A Solution containing 4.2 grams of an organic the molar mass of the organic
compound in Kb of acetone =1.71 k kg mol-1
Solution:
Mass = organic compound,
Mass of acetone,(Solvent).
W 2 =4.2 g
W 1=50g
ΔTb=1.8
Kb=1.71 k kg mol-1
If the molar mass of the solute is M, then
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23.
ADDITIONAL PROBLEMS ON SOLUTIONS
M=
1000 K b w 2 1000 1.71 4.2
-1
-1
=
gmol =79.8 g mol
50. 1.8
w1 VTb
42. Calculate the molality of a KCI solution in water such that the freezing point is
depressed by 2 K . (Kf for water = 1.86 K kg mol-1)
Solution: For KCI, the freezing point depression is given by, Tf =iK f m
This gives,
m=
Tf
iK f
=
2
-1
-1
mol kg =0.54mol kg
2 1.86
43. A decinormal solution of NaCI exerts an osmotic pressure of 4.6 atm at 300 K.
calculate its degree of dissociation. (R = 0.082 L atm K-1 mol-1)
Solution:
No. of moles of NaCI per litre of solution =0.1
Osmotic pressure, =4.6 atm
Temperature,
T=300 K
Had NaCI not dissociated then
normal =CRT =0.1 0.082 300atm
But
=2.46atm
obs = 4.6 atm
As per dissociation
Observedemagnitudeofacolligatuveproperty
i =
Normalmagnitudeofacolligativeproperty
pobs=
=
4.6atm
pnormal
2.46atm
= 1.87
For the the dissociation of an electrolyte producing n ions,
i 1
a
n1
1.87101
21
So,Prcentage dissociation =100
0.87
1
0.87
= 100 0.87 =87 %
44. The degree of dissociation of Ca (NO3)2in a dilute aqueous solution containing 7.0 g
of the per 100g of water at 100oC 70per cent. If the vapour pressure of water at 100oC
is 760 mm , them calculate the vapour pressure of the solution’
Solution: Each molecule of calcium nitrate, (Ca(NO3)2 given 3 ions on dissociation in
solution e.g.,
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24.
ADDITIONAL PROBLEMS ON SOLUTIONS
Ca(NO 3)2Ca
So,
2+
+ 2NO
3
n = 3. For electrolytes which-dissociate into ions in Solution,
i-1
=
n -1
70
i =
(ν - 1) +
(3 - 1) + 1 = 2.4
100
1=
Normal molar mass of Ca(N03)2 = (40 + 2 x 14 + 6 x 16) g/mol =164 g/mol
From the Raoult's law
or,
o
p -p
n2
=
o
i
p
For dilute solution, n2 << n1.So.
760 - P
760
2
=
/M
i
w
2
= /M
w
n1 +n 2
2.4 (7/164)
=
(1000/18)
2.4 7 18
= 0.0184
164 100
1 1
760- p = 760 x o.0184 mm Hg = 14 mm Hg
p = (760-14) mm Hg = 746 mm Hg
45. Calculate the amount of KCl which must be added to 1 kg of water so that
the freezing point is depressed by 2 K. (Kf (water) = 1.86 K kg mol-l)
Solution:
Mass of water, w1 = 1 kg
Mass of KCl Per kg of water.= w2g
Molar mass of KCI = (39 + 35.5) g mol-1 = 74.5 g mol-l
We know,
For KCI,
2 =
w
1
ΔTf=I Kf m =i Kf
w
f 2
/745
1
iK
n
i=2
T
w
2
745
=
2
74.5
2
f
iK
1.86
=
74.5
1.86
g = 40.05g
f
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25.
ADDITIONAL PROBLEMS ON SOLUTIONS
46. Calculate the amount of sodium chloride which must be added to 1000 mL of water
so that its freezing point is depressed by 0.744 K For water Kf =1.86 K/ M Assume
density of water to be one g mL-1
Mass of NaCI , w2=?
Volume of water =1000 mL
ΔTf= 0.744 K
Kf (Water) =1.86 K/m
Density of water= 1 g mL-1
So,
Mass of water, w1=1000 mL 1 g mL-1=1000 g =1 kg
w
i K f w2
According to the definition, = iK m = i k 2 / 58.5
T
58.5 w
1
f
f
f
=
w1
T 58.5 1
0.744 58.5
f
w =
=
= 11.7g
iK
2 1.86
2
f
So
Mass of NaCI required = 11.7 g
Solution:
47.
Calculate the amount of sodium chloride (electrolyte) which must be added to
one kilogram of water so that the freezing point is depressed by K Given : Kf for water =
1.86 K kg mol-1
Solution: Mass of sodium chloride. w2=?
Molar mass of sodium chloride. M2 =58.5 g/mol
Mass of water
w1=1 kg
Kf=1.86 K kg mol-1
For NaCI
I=2
So,
= i K m = iK
T
f
f
f
w2 M2
w1
=
2 1.86 w2
58.5 1
T
58.5
3 58.5
f
w =
=
g = 47.18g
or,
2 1.86
2 1.86
2
48.The freezing point depression. of o.1 m NaCl solution is 0.372 C. What conclusion
would you draw about its molecular state? Kf for water is 1.86 K kg mol-1
Solution: Molality ·of NaCl solution,
For ionic substances , we have
or
m = 0.1
ΔTf, =0.3 72 C
Kf = 1.86 K kg mol-1
ΔTf= i K1m
0.372 = i X 1.86 X 0.1
0.372
i=
= 2
1.86 0.1
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26.
ADDITIONAL PROBLEMS ON SOLUTIONS
The value of i=2, indicates that the solute NaCI in solution is completely dissociated
giving two ions, i.e., NaCI
Na+ (aq) + Cl-(aq).
49. The vapour pressure of a pure liquid A is 40 mm Hg at 310 K The vapour pressure
of this liquid in a solution with liquid B is 32 mm Hg . Calculate the molefraction of A in
the solution if it obeys the Raoult’s law.
o
Solution: Vapour pressure of pure A, P A = 40 mm Hg
Vapour pressure of A solution, PA = 32 mm Hg
o
According to the Raoult’s law,
PA = PA XA
PA
32 mm Hg
0.8
o
40 mm Hg
A
PA
50. A solution of 12.5 g urea in 170 g urea in 170 g of water gave a boiling point
elevation of 0.63 K Calculate the molar mass of urea, taking Kb=0.52 K/m.
Then,
X
=
=
Solution: Given: Mass of urea, of urea, w2=3.5 g
Mass of water, w1=170 g=0.17 kg
Elevation of boiling point, ΔTb=0.63 K
Molar mass of the solute (urea) is given by,
K
w
b
2 = 0.52 12.5
g/ mol = 60.7 mol-1
M=
w
VT
0.17 0.63
1
b
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