Mole concept numerical ..........

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

Mass of ethyl alcohol/ litre = =100g500

Mass in g / litre 100

No. of moles of alcohol/ litre = = = = 2.18Molar mass in g /mol 46

Thus, the solution contains 2.18 moles of ethyl alcohol per litre of the solution. So, themolarity 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:

Mass of H2S04 per litre = 9.8 gMolar mass of H2SO4 = 2 x (1) + 32 + (4 x 16) = 2 + 32 + 64 = 98 g mol-1

98Therefore, Number of moles of H2S04 per litre of solution = = 0.1

98Molarity of H2S04 in solution= 0.1 moll 1-1

Hence,

As there are two equivalents per mole of H2 SO4, thereforeNormality 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

Mg = 180 g mot-1Molar mass of glucose,

Mass of water, ww = 30 g

BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL


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,

2.28Number of moles of glucose m 30 g water = mol = 0.0.157 mol180

0.0157×1000 =. 0.522' mol kg-1So, 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 = = mol =· 0 0157 molMolar mass of glucose 180g

Mass of water 30n w = mol =

1.667molMolar mass of water 18

ng 0.0157

Therefore, Molefraction of glucose = = = 0.01ng + n w 0.0157 +1.667

n w 1.667

Molefraction of water = = = 0.99ng + n w 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= 320g 1Then, Mass of methyl alcohol in

200320g

So, No. of moles of alcohol per 1000 g of water = 10 mol32g/ mol

Thus, I kg of the solution contains 10 mole of methyl1 alcohol. So, the molality of thesolution is 10 mol /kg.



5. An agueous of solution of a. dibasic acid (molar mass=118 g/mol) containing 17.7 gof 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,

wM

= 17.7 g/L= 118 g mol-1

Mass of solute (g/L) wSo 'No ·of moles of solute dissolved = = mol/L

-1Molar mass (g mpl ) M

17.7

Then Mass of the solvent, (water)

Volume of the solution Density of solution Mass of solvent (water)

= Mol /L = 0.15 mol /L118

= 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

So,

Then, Maas of the solvent, (water)

Moles of solute 990N o. of moles of solvent = = = 55.0

Molar mass in g / mol

18Total number of moles in solution = Moles of solute + Moles of solvent

= 0.15 +55 = 55.15Since, 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 1Mass of solute (acid) per 1000 g of solvent =17.7 × × = 0.151mol /

kg990 g 118

Mass in g per 1000 g solventNo. of moles of the solute/ 1000 g of solvent =

Molar mass (g/ mol)

1000 1=17.7 × × = 0.151 mol / kg

990 118

So, Molality = 0.151 mol/ kg



nacid 0.15 0.15

(iii) Mole- fraction (X) = = = = 0.027n n

acid+ water

0.15 + 55

55.15

6. How much sodium chloride be dissolved to make l litre of 0.1 F solution?

Solution:.NaCl. Therefore, So,

Sodium chloride is an ionic .compound and is represented by the formula

its gram-formula mass is, 23 g + 35.5 g1 litre of 1 F NaCl solution, one requires1 litre of 0.1 F NaCI solution, one. requires

= 58.5, g= 58.5 g, of NaCl= 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 Msolution? Atomic masses are; K= 39 u, Cl-= 35.5 u

Solution: Potassium chloride (KCl) is an ionic compound with a molar mass of39 + 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 × 0.1For 1 litre of 0.1 M KCl solution, one requires = 7.45g

1Thus, one requires 7.45 g of KCl to prepare 1 litre of its 0.1 M solution8. 2.46 g of sodium hydroxide (molar mass = 40 g/mol) are dissolved in water and thesolution 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 · solutionMolarity of the solution

= 100 mL=?

Molarity of a solution is given by the number of moles of solute present in 1000 mL ofthe solution, i.e.,

No. of moles of solute x 1000 ml L-1Molarity =Volume of the solution in mL



(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 -1×10molL- = 0.615molL= molL -1=

100 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

Solution: Molar mass of sodium nitrate, (NaN03) = (23 + 14 + 48) g/mol = 8S

g/molMass of 1dm3 (or 1 litre) of the solution = Volume x Density Therefore,

= 1000 cm3 x1.25 g/cm3 = 1250 gTherefore,Mass of water containing 8.5 g .of NaN03 = (1250- 85) g = 1165 g = 1.165 kgSo, 1mol -1Molality (m) of the solution. = = 0.86 mol

kg1.165kg

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 dissolvedVolume of solution

= 0.75g= 250 mL

L-Molar mass of sodium bicarbonate . = (23+ 1 + 12+~8) g mol-1 = 0.615

1 g mot'Since, one molecule of NaHCO3 contains only one cationic charge (on Na+), hence

Molar mass 84= g equiv

-1Equivalent mass, (E)=

1 1As per definition,

(Mass of NaHCO3 / Molar mass of NaHC03 )Molarity (M) = x 1000

Volume of solution in mL

-10.75g/84 mol 0.75/84mol -1= 0.0357molL= ×1000mL×

250/1000L -1

250 mL



(Mass of NaHCO3 / Equiv. mass of NaHC03 )

× 1000 ML / LNormality (N) =Volume of solution in mL

11. Calculate the molefraction of water in a mixture of 12 g water .108 g acetic acid and92 g ethyl alcohol.

Solution: Following the procedure of the previous problem, one can writeMass of water 12

n(H2O) = = = 0.67Molar mass

Mass of ethanol18

92n(C2H5OH) = = = 2.00

Molar mass 46

Mass of acetic acid 108n(CH3COOH) = = =1.80

Molar mass 60

So,Total number of moles in the solution, ntotat = 0.67 + 2.00 + 1.80 = 4.47

nH2O

0.67

Therefore, X water = = = 0.15

ntotal 4.47

nC2H5OH

2.00

Xethanol = = = 0.45ntotal 4.47

nCH3cooH

1.80

Xacetic acid = = = 0.40ntotal 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?

Solution: Mass of ammonia w2 = 0.85 g

M2=17 g mol-1

w1= 100 cm3 0.85g cm-3=0.85 g

Molar mass of ammonia,

Mass of liquid (solvent),



nNH w /M -13 2 2= mol kgSo, Molality of ammonia =Mass of solvent in kg 85/10000.85/17

85/1000

mol kg-1 = 0.59 mol kg-1or, Molality of ammonia =

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

= 2 × 12 + 2 × 1 + 4 × 16 + 2 2 +

16

= 24 + 2 + 64 + 36 = 126 g mol

Molar mass M of dibasic acid

-1

So,

Molar mass 126 -1Equivalent mass of the acid = = = 63 g equiv2 2

According to the definition,

Mass of solute in g per L

1.26 g / L -1Molarity = = = 0.01mol LMolar mass 126 g/mol

Mass of solute in g per L

1.26 g / L

-1and, Normality = =-1

= 0.01mol LEquivalent mass 63 g equiv

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:

= 100 cm2

=1.10 g /cm3

Volume of the solution

Density of the solution

Therefore 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



15gThen, mass H2SO4 in 110 g (=100 cm3) of solution = × 110g =

16.5g100g

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

100 × 16.5gMass of H2SO4 (95%) required = = 17.37g

95

Density of 95% H2SO4 =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) unitsDensity 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

106 ppm = 5.8 ppmSo, Concentration of dissolved oxygen =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

=1000 cm3

=0.85 g cm3

=Volume x Density= (1000 x 0.85) g= 850 g

Volume of the solution


So, Mass of 1000 cm3 of solution

Therefore Mass of solvent = (850-2.5) = 847.5 g

Thus 847.5 g of solvent contains 2.5 g of solute



2.5Or, 847.5 g of solvent contains mol of solute

1.38

2.5 11 g solvent contains = × mol

1.38 847.5

1000 2.51000 g of solvent contains = × = 0.0214 mol

8475 138

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

(c) 100 mL of 0.5 M H2SO4

(b) 2 L of N /10 HCI solution

(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,

106= 53 g equiv -1Equivalent mass of Na2 CO3= =

2

Thus 1 L of 1 N Na2CO3 solution should contain 53 g of Na2 CO3

53Then, 1 L of N/ 10 Na2CO3 solution would contain g of Na2 Co3 = 5.3 g Na2CO3= 5.3

10g 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,



36.5 × 2 732. L of N /10 HCl solution would contain = g of HCl = g HCl =7.3 g HCl

10 10

(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

98 0.5Then, 100 mL of 0.5 M H2SO4 would contain = × 100 × g of H2SO4

1000 1

= 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 = (212+416 +2 1) +2 (2+1 +16)

=(24 +64 + 2 +36) =126 g mol-1

Basicity of oxalic acid =2

So,

126= 63 g equiv -1Equivalent mass of oxalic acid =

2

Then, 1 L of 1 N solution of oxalic acid contains 63 g oxalic acid

1000 mL of 1 N “ “ “ “ 63 g oxalic acid

N 63 1× 250 × g oxalic

250 mL of ” “ “ “ = acid10 1000 10

63= = 1.575 g oxalic acid

1000

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



Mass of solution =100 g

=1.10 g cm-3

=98 g mol -1


Molar mass of the H2SO4

So, Mass of water containing 15 g of H2SO4 = (100 – 15 )g =85 g

15 g 15No. of moles of H2SO in 15 g of H2SO = = mol = 0.153

mol98 g/ mol 98

mass=

Density

100g 90.93 3and, Volume of 100 g of solution = = 90.9 cm dm1000-31.1g cm

(i) Calculation of molarity

No. of moles of H2SO4 0.153 mol -3Molarity of H2SO4 solution = = = 1.68 mol dm

3 3Volume of solution in dm (90.9 / 1000) bm

So, thee molarity of H2SO4 solution is 1.68

(ii) Calculation of molality

No. of moles of H SO42Molarity of H2SO4 solution = 1000Mass of water in g0.153 -1 -1Molality = = × 1000mol kg = 1.8 mol

kg85

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

=60g mol-1Molar mass of urea (NH2CONH2)


ADDITIONAL PROBLEMSnurea

ON SOLUTIONS

10/60 1.1667X = = =urea

nurea + n (10/60 + 90/18 0.1667 + 5

Then, Mole- fraction of urea, water

0.1667=

5.1667Xurea = 0.032

po o- p Then, from Raoult’s law, = Xurea

po

55.3 - Ps = 0.03255.3

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.

PoSolution Given Vapour pressure of water (solvent) =18.15 torrA

Vapour pressure of solution, PA=18.06 torr

WB = 9.47 g

WA=100g

Mass of sugar (solute)

Molar of water (solvent)

Molar mass of water (solvent) MA=18 g /mol

Molar mass of sugar(solute), MB = ?

From Eq (3.11),

9.47×18

WB ×MA PA 18.03

MB = = g / mol =342 g/ molo

W 100 18.15 -18.06P -P A A A

oPW ×M 9.47×18 B A A 18.03

Form Eq. (3.13) M = = oP 18.15 -18.06

W 100-P B A A A



21 The vapour pressure of a 5% aqueous solution of a non – volatice organic substanceat 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

W1=(100-5) g= 95 g=0.095 kgMass of solvent (water)

Vapour pressure of solution, Ps =745 mm Hg

Vapour pressure of pure solution.(water) at373 K, Po =760 mm Hgs

Molar mass of solution, M2=?

Molar mass of solvent (water), M1 =18 g/ mol

oP - Ps n2 W2 / M2s

Form Raoult,s law, = =o

P W

+

W

n + n / M / Ms 1 2 1 1 2 2

5/M2

760 - 745

5/M2= =

5.278

+5/M

o95 /18+ 5/MP 760s 2 2

The above equation may be rewritten as

760 5.278 +5 / M2

5.278 M2

= = +1=1.0556M2 +1

5 /

M

15 52

50.67 =1.0556 M2 + 1

-150.67 -1

or M = 1.0556 = 47g mol1.05562

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.



= 1 molar = 1 mol L-1

= 1.03 · g mL-1

= (23 + 35.5) g/mol = 58.5 g/ mol-1

Solution. Conc. of the solutionDensity of solutionMolar mass of NaCl

So, 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

1×1000

Thus, Molality, of the solution, m = mol/kg = 1.0293 mol/kg971.5

Then, 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

So,

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

w2 =2.4 g

w1=100g =0.1 kg

ΔTb=0.21 K

Solution: Mass of solute.

Mass of solvent,

Molar mass of the solute=? (M2)

K b × w 2 0.52× 2.4

we know, M2= = g/mol = 59.4 g/ molw1× Tb 0.1×0.21

24. A solution contains 3.5 g of a non-volatile solute in 125 g of water; and it boils at373.52 K. Calculate the molar mass of the solute. (Kb for water= 0.52 K/m)Solution· Mass· of the solute,

Mass of water,Elevation of boiling point,

w2 = 3:5 gw1 =125 g = 0.125 kgΔ Tb = (373.52- 373.0) K = 0.52 K.

= M (?)Let, Molar mass of the solute

We know that,



K

w

/ M 0.52 kg mol-

1×3.5g

bK b n2 2 Tb = =

w1 w1 M ×0.125 kg

-10.52×3.5 K kg mol g

0.52 K=M ×0.125 kg

-10.52×3.5 K kg mol g 3.5 -1

g mol = 28 g mol-1

M = =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 isdissolved 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 mLTaking 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

K b w 2We know, ΔTb = ×

w1K b w

2

M

0.6×1.5

or M= × g/ mol = 17.3 g/ mol Tb w1 0.52×0.1

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 gMass of carbon disulphide (solvent) = 100 g = 0.1 kg

Molar mass of sulphur = MΔTb = (46.66 - 46.30)°C = 0.36°C

K b w 2 2.42×3.795We know, M= = g/mol =

255.1g/molw1 Tb 0.1×0.36



Atomic mass of sulphur = 32 g/molSo,

255.1No. of sulphur atoms in one molecule = = 7.97 =

832

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 and0.5 12 deg kg mol-1 respectively?Solution:

We have the relationships,

K b .n2 K f .n2

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 Kf f

Tf ×K b 0.4×0.512 ooC = 0.11

C

or Tb = =K f 1.86

Boiling point of the urea solution = 100o C + 0.11oC = 100. 11oCThen,

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

Concentration of solution=278.82 K=0.25 molal =0.25 mol/kg

Freezing point depression Tf = 2K

K f ?

Tf =K f ×mWe know that,



-1 Tf 2K

K f = = 8 K kg molm 0.25 mol/kg

29. Find the (i) boiling point, and (ii) freezing point of a solution containing 0.520 gglucose (C6H12 O6) dissolved in 80.2 g of water for water. Kf =1.86 K/m, and Kb =0.52K/m.

Normal boiling point of water = 100oC Normal freezing point of water = 0.0oC

Solution:

Mass of glucose = 0.52 gMolar mass of glucose (C6H12O6 =180 mol

80.2Mass of water =80.2 g = kg

1000K n

b 2We know T =

b

where, n is no. of moles of solute (w /2 2wM2) w1 is the mass of solvent in kg1

0.52 × 0.520.52×(w2/M2 )Tb = 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 × 0.522 2fSimilarly T =

f= 0.067 K

180 × (80.2/1000)w1

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-1Solution: Mass of water (solvent) w1=1 kg

o Tf =10 C

K f =1.86 K kg

mol

w2=?

K f (w 2 / 62)

-1

Mass of ethylene glycol required,

We know, Tf

=w1



1.86× w 2

10 =62×1

10×62×1

This gives w = g = 333.3g1.862

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 pressureTemperature, Mass of solute, Volume of solution,

= (5000/ 76) atmT = (27 +273) =300 K W= 4 gV=100 cm3 atm deg-1 mol-1

R = 0.0821 dm3atm deg-1 mol-1

wWe know, pv

=RT

M

w 4×0.0821×300

500 / 76 × 0.1

This gives, M = × RT g/mol = 149.7g/molV

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 soluteSolution: Mass of solute,

Volume of solution, Osmotic pressure Temperature,Molar mass of solute

w = 5 g = 0.005 kgV= 250 cm3 =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

-1 -1wRT 0.005 kg × 8.314 JK m,ol × 298

KM = =We know, 5 -2 -4 3V 4 × 10 N m × 2.5 × 10 m

M=0.1239 kg mol-1 = 123.9 g mol-1or



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,

Volume of the solution,w=7.0 gV = 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

we have,

atm760

wRT 7.0×0.0821×310 7.0×0.0821×310×760M= = = =67700 g / mol

pV (20/760)×0.1 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 KV=100 cm3 =0.1 L W= 3.02 g=2.55 atm

Volume of the solutionMass of Solute, Osmotic pressure,

w

The osmotic pressure is given by, V=n RT = RT M

wRT 3.02×0.0821×298

So. M = = = 67700 g / molpV 2.55×0.1

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 gRelative molar of sugar = (12 12+22+1116) =144+22+176=342We know,

nv = nRT



n w RTor, = RT = ×

V M V

5 RT Sugar =So, ×

342 V

0.877 RTand, A = ×

M VSince, the two solutions are isotonic, hence, Sugar = A, So,

0.877 RT 5 RT

× = ×M V 342 V

0.877×342

or, M = = 60g/mol5

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).

Kb ×n2Solution: We know, Tb

=where n2 is the no. of moles of solute; w1 is

w1the mass of the solvent in kg.

Mass of solute

1.0 gn2 = =

Molar mass of the solute M

and w1=100g =0.1kg

Substituting these values, one can write,

1.84K kg mol-1 1.0g

0.307K = ×0.1Kg M

1.84K kg mol-1

×1.0 1.84×1.0gSo, M = × g/ mol= 59.9 g/ mol

0.1Kg×0.307 K 0.1×0.30737. 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,



Mass of 1 cm3 of water =0.997 gTherefore, 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 55.39Therefore,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 bymass. Calculate the molefraction of each component.Solution: Let us consider 100 g of the solution .Then,

Mass of water,

Mass of ethanol,

Mass of acetic acid

mw=25 g

meth =25g

macet=50 g

Molar mass of water, ethanol and acid are, 18 g/ mol, 46 g / mol, and 60 g/

mol respectively Then,

25gNo. of moles of water, nw= = 1.3918 g/mol

25g

No. of moles of ethanol, n eth= = 0.5446g/mol

50gNo. of moles of acetic acid n acet= =0.83

60g/mol=(1.39 +0.54+0.83) =2.76

1.39

Total no of moles in solution

So, Mole –fraction of water, Xw= = 0.502.76

0.54Mole- fraction of ethanol, Xeth = =0.20

2.76

0.83Mole- fraction of acetic acid Xacet = =0.30

2.76

39. Concentrated sulphuric acid has a density of 1.9 g/ ML and is 99% H2SO4byweight 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 gSince, the given acid is 99% pure, hence



190099Actual mass of H2SO4 in 1 litre sample = = 1881g

100Molar mass of H2SO4=(2+32+64)g mol-1=98 g mol-1

1881g

So, No of moles of H2SO4 in 1 litre of the sample = = 19.19 mol-198 g mol

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 gMass of HNO3 in 100 g of sample = 31 g

Density of conc. HNO3 sample =1.41 g cm-3

Mass 2100gSo, Volume of 100 g of the HNO3 sample = = = 70.92cmcm-3Dwnsity 1.41gThus, 69 g of HNO3 is contained in .70.92 cm3 of conc. HNO3

70.92

1g “ “ “69

70.92 323g “ “ 23 “ =23.6 cm69

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).

W2 =4.2 gW1=50g

ΔTb=1.8Kb=1.71 k kg mol-1

If the molar mass of the solute is M, then



1000×Kb × w 2 1000×1.71×4.2 -1 -1M= = gmol =79.8 g mol

w1× VTb 50.×1.8

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

Tf 2 -1 -1This gives, m= = mol kg =0.54mol

kgiK f 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 atmTemperature, T=300 K

Had NaCI not dissociated then

normal =CRT =0.1× 0.082 × 300atm =2.46atm

But obs = 4.6 atm

As per dissociation

Observedemagnitudeofacolligatuveproperty pobs= 4.6atm

i = = = 1.87pnormalNormalmagnitudeofacolligativeproperty 2.46atm

For the the dissociation of an electrolyte producing n ions,

i 1 1.87101 0.87

a 0.87n1 21 1

So,Prcentage dissociation =100 = 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 100oCis 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.,



Ca(NO )2 Ca2+

+ 2NO -3 3So, n = 3. For electrolytes which-dissociate into ions in Solution,

i -1

=n -1

70

i = (ν - 1) + 1=or, × (3 - 1) + 1 = 2.4100

Normal molar mass of Ca(N03)2 = (40 + 2 x 14 + 6 x 16) g/mol =164 g/mol

From the Raoult's law

op -p n2

= io n +n1 2pFor dilute solution, n2 << n1.So.

iw /M 760 - P

2

2.4 × (7/164) 2.4 × 7 × 18

2= = = = 0.0184760 w /M

1 1(1000/18) 164 ×

100

760- p = 760 x o.0184 mm Hg = 14 mm Hgp = (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.= w2gMolar mass of KCI = (39 + 35.5) g mol-1 = 74.5 g mol-l

iK × w /745n

f 22We know, ΔTf=I Kf m =i Kf =w 1

1For KCI, i=2

T × 745f 2 × 74.5 74.5

g =w = = 40.05giK 2 × 1.86 1.862

f



46. Calculate the amount of sodium chloride which must be added to 1000 mL of waterso 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

Solution: Mass of NaCI , w2=?

Volume of water =1000 mLΔTf= 0.744 K

Kf (Water) =1.86 K/mDensity of water= 1 g mL-1

Mass of water, w1=1000 mL 1 g mL-1=1000 g =1 kg

So, w2 / 58.5

× =

i Kf w2According to the definition, T = iK m = i k

w1

58.5 × w1f ffT × 58.5 ×

1 f 0.744 × 58.5w = = = 11.7g

iK 2 × 1.86

2f

So Mass of NaCI required = 11.7 g

47. Calculate the amount of sodium chloride (electrolyte) which must be added toone 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.Mass of water

M2 =58.5 g/mol

w1=1 kgKf=1.86 K kg mol-1

I = 2For NaCI

w2 × M2

2 × 1.86 × w2So, T

f= i K m = iK × =

w1 58.5 × 1f f

Tf

× 58.53 × 58.5

or, w = = g = 47.18g2 1.86 2 × 1.862

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, m = 0.1ΔTf, =0.3 72°CKf = 1.86 K kg mol-1

ΔTf= i K1m

0.372 = i X 1.86 X 0.10.372

For ionic substances , we have

or i = = 21.86 × 0.1



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 pressureof 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.

oSolution: Vapour pressure of pure A, P A = 40 mm HgVapour pressure of A solution, PA = 32 mm Hg

oAccording to the Raoult’s law, PA = P A XA

PA 32 mm HgThen, X = = 0.8

oP A

40 mm HgA

50. A solution of 12.5 g urea in 170 g urea in 170 g of water gave a boiling pointelevation of 0.63 K Calculate the molar mass of urea, taking Kb=0.52 K/m.Solution: Given: Mass of urea, of urea, w2=3.5 g

Mass of water, w1=170 g=0.17 kgElevation of boiling point, ΔTb=0.63 K

Molar mass of the solute (urea) is given by,

K × w0.52 × 12.5b 2 g/ mol = 60.7 mol-1M = =

w × VT 0.17 × 0.631 b


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