Solved Examples for JEE: Solutions | Chemistry for JEE Main & Advanced PDF Download

Example 1. The molarity of 20% (W/W) solution of sulphuric acid is 2.55 M. The density of the solution is: 
(a) 1.25 g cm^-3
(b) 0.125 g L^-1
(c) 2.55 g cm^-3
(d) unpredictable
Ans. (a)
Solution.
Volume of 100 g of solution = 100/d ml

= 1.249 ≈ 1.25

Example 2. The density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Calculate the molarity and normality of the solution-
(a) 1.445 M
(b) 14.45 M
(c) 144.5 M
(d) 0.1445 M
Ans. (a)
Solution.

Volume of 100 gram of the solution = 100/d


Number of moles of H₂SO₄ in 100 gram of the solution = 13/98


= 1.445 M

Example 3. Calculate the molarity of pure water (d = 1g/L)
(a) 555 M
(b) 5.55 M
(c) 55.5 M
(d) None
Ans. (c)
Solution.
Consider 1000 mL of water
Mass of 1000 mL of water
= 1000 × 1 = 1000 gram
Number of moles of water = 100/18
= 55.5

= 55.5/1
= 55.5 M

Example 4. Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml of 0.1 M solution-
(a) 2.65 gram
(b) 4.95 gram
(c) 6.25 gram
(d) None
Ans. (a)
Solution.
We know that

where;
W = Mass of Na₂CO₃ in gram
M = Molecular mass of Na₂CO₃ in grams = 106
V = Volume of solution in litres = 250/1000 = 0.25
Molarity = 1/10
Hence, = 1/10

or

= 2.65 gram

Example 5. The freezing point of a 0.08 molal aqueous solution of NaHSO₄ is -0.372°C. The dissociation constant for the reaction
HSO₄^- ⇌ HSO₄^2-is
(a) 4 × 10^-2 
(b) 8 × 10^-2 
(c) 2 × 10^-2 
(d) 1.86 × 10^-2 
Ans. (a)
Solution.
ΔT_f = K_f × m

This means that total molal conc. of all particles is 0.2.
NaHSO₄ → Na⁺ + HSO₄^-
0.08                         0.08
HSO₄^- ⇌  H⁺ + SO₄^2-
(0.08 - x)       x           x
The net particle concentration
0.087 = (0.08 - x) + x + x = 0.2
or x = 0.04


Example 6. Find the molality of H₂SO₄ solution whose specific gravity is 1.98 g ml^-1 and 95% by volume H₂SO₄
(a) 7.412 
(b) 8.412 
(c) 9.412 
(d) 10.412
Ans. (c)
Solution.
H₂SO₄ is 95% by volume
wt. of H₂SO₄ = 95g
Vol of solution = 100ml
Moles of H₂SO₄ = , and weight of solution = 100 × 1.98 = 198 g
Weight of water = 198 - 95 = 103 g

= 9.412
Hence molality of H₂SO₄ solution is 9.412

Example 7. Calculate molality of 1 litre solution of 93% H₂SO₄ by volume. The density of solution is 1.84 gm ml^-1
(a) 9.42 
(b) 10.42 
(c) 11.42 
(d) 12.42 
Ans. (b)
Solution.
Given H₂SO₄ is 93% by volume
wt. of H₂SO₄ = 93g
Volume of solution = 100ml
weight of solution = 100 × 1.84 gm
= 184 gm
wt. of water = 184 - 93 = 91 gm


= 10.42

Example 8. Suppose 5gm of CH₃COOH is dissolved in one litre of Ethanol. Assume no reaction between them. Calculate molality of resulting solution if density of Ethanol is 0.789 gm/ml.
(a) 0.0856 
(b) 0.0956 
(c) 0.1056 
(d) 0.1156 
Ans. (c)
Solution.

Fig: Acetic acid structure

Wt. of CH₃COOH dissolved = 5g
Eq. of CH₃COOH dissolved = 5/60
Volume of ethanol = 1 litre = 1000ml.
Weight of ethanol = 1000 × 0.789 = 789g
Molality of solution

 
= 0.1056

Example 9. Calculate the molarity and normality of a solution containing 0.5 gm of NaOH dissolved in 500 ml. solution-
(a) 0.0025 M, 0.025 N 
(b) 0.025 M, 0.025 N
(c) 0.25 M, 0.25 N 
(d) 0.025 M, 0.0025 N  
Ans. (b)
Solution.
Wt. of NaOH dissolved = 0.5 gm
Vol. of NaOH solution = 500 ml
Calculation of molarity 
0.5 g of NaOH = 0.5/40 moles of NaOH
[∴ Mol. wt of NaOH = 40]
= 0.0125 moles
Thus 500 ml of the solution contain NaOH = 0.0125 moles
1000 ml of the solution contain

= 0.025 M
Hence molarity of the solution = 0.025 M
Calculation of normality 
Since NaOH is monoacidic;
Eq. wt. of NaOH = Mol. wt. of NaOH = 40
0.5 gm of NaOH = 0.5/40 gm equivalents = 0.0125 gm equivalents
Thus 500 ml of the solution contain NaOH = 0.0125 gm equ.
1000 ml of the solution contain

Hence normality of the solution = 0.025 N

Example 10. Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 gm of urea per 250 gm of water (Mol. wt. of urea = 60).
(a) 0.2 m, 0.00357 
(b) 0.4 m, 0.00357 
(c) 0.5 m, 0.00357 
(d) 0.7m, 0.00357
Ans. (a)
Solution.
Wt. of solute (urea) dissolved = 3.0 gm
Wt. of the solvent (water) = 250 gm
Mol. wt. of the solute = 60
3.0 gm of the solute = 3.0/60 moles = 0.05 moles
Thus 250 gm of the solvent contain = 0.05 moles of solute
1000 gm of the solvent contain

= 0.2 moles
Hence molality of the solution = 0.2 m
In short,
Molality = No. of moles of solute/1000 g of solvent

= 0.2 m
Calculation of mole fraction 
3.0 gm of solute = 3/60 moles = 0.05 moles
250 gm of water = 250/18 moles = 13.94 moles
Mole fraction of the solute

= 0.00357 

Example 11. A solution has 25% of water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component.
(a) 0.50, 0.3, 0.19 
(b) 0.19, 0.3, 0.50 
(c) 0.3, 0.19, 0.50 
(d) 0.50, 0.19, 0.3
Ans. (d)
Solution.
Since 18 g of water = 1 mole
25 g of water = 25/18 = 1.38 mole
Similarly, 46 g of ethanol = 1 mole
25 g of ethanol = 25/46 = 0.55 moles
Again, 60 g of acetic acid = 1 mole
50 g of acetic acid = 50/60 = 0.83 mole
Mole fraction of water

= 0.50
Similarly, Mole fraction of ethanol

= 0.19
Mole fraction of acetic acid

= 0.3

Example 12. 15 gram of methyl alcohol is dissolved in 35 gram of water. What is the mass percentage of methyl alcohol in solution?
(a) 30% 
(b) 50% 
(c) 70% 
(d) 75%
Ans. (a)
Solution.
Total mass of solution = (15 35) gram = 50 gram
mass percentage of methyl alcohol

= 30% 

Example 13. Calculate the masses of cane sugar and water required to prepare 250 gram of 25% cane sugar solution?
(a) 187.5 gram, 62.5 gram 
(b) 62.5 gram, 187.5 gram
(c) 162.5 gram, 87.5 gram 
(d) None of these
Ans. (b)
Solution.
Mass percentage of sugar = 25
We know that
Mass percentage

or Mass of cane sugar

= 62.5 gram
Mass of water = (250 - 62.5) = 187.5 gram 

Example 14. The density of a 3M sodium thiosulphate solution (Na₂S₂O₃) is 1.25 g/mL. Calculate
(i) the percentage by mass of sodium thiosulphate, 
(ii) the mole fraction of sodium thiosulphate and 
(iii) molalities of Na  and S₂O₃²-ions -
Solution. 
(i) Mass of 1000 mL of Na₂S₂O₃ solution
= 1.25 × 1000 = 1250 g
Mass of Na₂S₂O₃ in 1000 mL of 3M solution
= 3 × Mol. mass of Na₂S₂O₃
= 3 × 158 = 474 g
Mass percentage of Na₂S₂O₃ in solution

Alternatively, 

(ii) No. of moles of Na₂S₂O₃ 

Mass of water = (1250 - 474) = 776 g
No. of moles of water

Mole fraction of Na₂S₂O₃

(iii) No. of moles of Na⁺  ions
= 2 × No. of moles of Na₂S₂O₃
= 2 × 3 = 6
Molality of Na  ions

No. of moles of S₂ ions = No. of moles of Na₂S₂O₃ = 3
Molality of ions 

Example 15. A solution is perpared by dissolving 5.64 gm of glucose in 60g of water. Calculate the following.
(i) mass per cent of each of glucose and water,
(ii) molality of the solution,
(iii) mole fraction of each of glucose and water.
Solution. 
(i) Total mass of solution
= 5.64 60 = 65.64 g
Mass per cent of glucose 

Mass per cent of water
= (100 -Mass per cent of glucose)
= (100 -8.59) = 91.41%
(ii) No. of moles of glucose
=

Mass of water in kg
=

(iii) No. of moles of glucose 

No. of moles of water 

Mole fraction of glucose

Mole fraction of water

Example 16. The volume of water which must be added to a mixture of 350 cm³ of 6 M HCl and 650 ml of 3 M HCl to get a resulting solution of 3 M concentration is [Ans. D]
(a) 75 mL 
(b) 150 mL 
(c) 100 mL 
(d) 350 mL
Ans. (d)
Solution. 
Molarity of mixture of 6 M and 3 M HCl

Now, apply dilution formula

Volume of water to be added
= 1350 - 1000 = 350 ml

Example 17. The mole fraction of CH₃OH in an aqueous solution is 0.02 and its density is 0.994 g cm^-3. Determine its molarity and molality.
Solution.  Let x mole of CH₃OH and y mole of water be present in solution.
Mole fraction of CH₃OH 

So,

Example 18. Calculate the concentration of NaOH solution in g/mL which has the same normality as that of a solution of HCl of concentration 0.04 g/mL. -
Solution.

 
N_NaOH º N_HCl

Example 19. How many Na⁺ ions are present in 50 mL of a 0.5 M solution of NaCl ?
Solution. 
Number of moles of NaCl 

NaCl → Na⁺ + Cl¯
Number of moles Na  = Number of moles of NaCl = 0.025
Number of ions of Na  = 0.025 × 6.023 × 10²³ = 1.505 × 10²²

Example 20. 250 mL of a Na₂CO₃ solution contains 2.65 g of Na₂CO₃. 10 mL of this solution is added to x mL of water to obtain 0.001 M Na₂CO₃ solution. The value of x is :
(Molecular mass of Na₂CO₃ = 10⁶ amu)-
(a) 1000 
(b) 990 
(c) 9990 
(d) 90
Ans. (b)
Solution.  
Molarity of solution

M₁V₁ = M₂V₂
0.1 × 10 = 0.001 (10 x)
x = 990 mL

Example 21. The volumes of two HCl solutions A (0.5 N) and B (0.1 N) to be mixed for preparing 2 L of 0.2 N HCl are -
(a) 0.5 L of A 1.5 L of B       
(b) 1.5 L of A 0.5 L of B
(c) 1 L of A  1 L of B              
(d) 0.75 L of A 1.25 L of B
Ans. (a)
Solution.  
Let x L of A and (2 -x) L of B are mixed.
M₁V₁  M₂V₂ = M_R (V₁ V₂)
0.5 × x 0.1 (2 -x) = 0.2 × 2
(0.5 -0.1) x = 0.4 -0.2
0.4 x = 0.2
x = 0.5 L
0.5 L of A and 1.5 L of B should be mixed.

Example 22. Lowering of vapour pressure due to a solute in 1 molal aqueous solution at 100ºC is -
(a) 13.44 mm Hg   
(b) 14.12 mm Hg   
(c) 31.2 mm Hg   
(d) 35.2 mm Hg 
Solution.  

x_A = 1 -0.0176 = 0.9824
p = p°x_A
= 760 × 0.9824 = 746.62
Dp = p° - p = 760 -746.62
= 13.4

Example 23. The mass of a non-volatile solute (molecular mass = 40) which should be dissolved in 114 g octane to reduce its vapour pressure to 80% will be -
(a) 20 g   
(b) 30 g   
(c) 10 g   
(d) 40 g
Ans. (c)
Solution. 
If p° = 100, then p = 80
p = p° x_A
80 = 100 × x_A 
x_A = 0.80

Example 24. A solution of urea in water has boiling point of 100.15ºC. Calculate the freezing point of the same solution if K_f and K_b for water are 1.87 K kg mol^-1 and 0.52 K kg mol^-1 respectively -
Solution. 
ΔT_b = (100.15 -100) = 0.15ºC
We know that, ΔT_b = molality × K_b

ΔT_f = molality × K_f
= 0.2884 × 1.87 = 0.54ºC
Thus, the freezing point of the solution
= -0.54ºC.

Example 25. Calculate the molal depression constant of a solvent which has freezing point 16.6ºC and latent heat of fusion 180.75 J g^-1 -
Solution. 

R = 8.314 J K^-1 mol^-1,
T_f = 16.6ºC = 273 16.6 = 289.6 K,
L_f = 180.75 J g^-1
Substituting the values in the above equation,

Example 26. The freezing point depression of 0.001 m K_x[Fe(CN)₆ ] is 7.10 × 10^-3 K. Determine the value of x. Given, K_f = 1.86 K kg mol^-1 for water -
Solution. Dx = i × K_f × m
7.10 × 10^-3 = i × 1.86 × 0.001
i = 3.817

Molecular formula of the compound is
K₃ [Fe(CN)₆]

Example 27. A certain substance `A' tetramerises in water to the extent of 80%. A solution of 2.5 g of A in 100 g of water lowers the freezing point by 0.3ºC. The molar mass of A is -
(a) 122       
(b) 31      
(c) 244     
(d) 62
Ans. (d)
Solution. 

ΔT = iK_f × m

Example 28. van't Hoff factor of Hg₂Cl₂ in its aqueous solution will be (Hg₂Cl₂ is 80% ionized in the solution) -
(a) 1.6    
(b) 2.6     
(c) 3.6    
(d) 4.6 
Ans. (b)
Solution. 

Example 29. Calculate the amount of NaCl which must be added to 100 g water so that freezing point is depressed by 2K. For water, K_f = 1.86 K kg mol^-1 -
Solution. NaCl is a strong electrolyte. It is completely dissociated in solution.
Degree of dissociation, a = 1
NaCl  Na⁺¹  Cl^-1
(n = 2)
No. of particles after dissociation = 1 (n -1) a
= 1 (2 -1) × 1 = 2

Let w g of NaCl be dissolved in 100 g of water.
So, 

Example 30. The degree of dissociation of Ca(NO₃)₂ in a dilute solution containing 14 g of the salt per 200 g of water at 100ºC is 70%. If the vapour pressure of water is 760 mm, calculate the vapour pressure of solution -
Solution. Dp_theo. = Lowering in vapour pressure when there is no dissociation

(given, p° = 760 mm, w = 14 g, W = 200 g, M = 18, m = 164)

Degree of dissociation 

Ca(NO₃)₂ Ca²⁺ + 2NO₃^-
(n = 3)

So, ΔT_obs. = 2.4 × ΔT_theo. = 2.4 × 5.84
= 14.02 mm
p -p_s = Dp_obs. = 14.02
p_s = p_s -14.02 = 760 - 14.02 = 745.98 mm

Example 31. Calculate the normal boiling point of a sample of sea water found to contain 3.5% of NaCl and 0.13% of MgCl₂ by mass. The normal boiling point of water is 100ºC and K_b (water) = 0.51 K kg mol^-1. Assume that both the salts are completely ionised -
Solution. 
Mass of NaCl = 3.5 g
No. of moles of NaCl

Number of ions furnished by one molecule of NaCl is 2.
So, actual number of moles of particles furnished by sodium chloride = 2 × Similarly, actual number of moles of particles furnished by magnesium chloride
= 3 × Total number of moles of particles 

Mass of water = (100 -3.5 -0.13) = 96.37 g 

ΔT_b = Molality × K_b
= 1.2846 × 0.51 = 0.655 K
Hence, boiling point of sea water = 373.655 K or 100.655ºC. 

Example 32. Sea water is 3.5% by mass of a salt and has a density 1.04 g cm^-3 at 293 K. Assuming the salt to be sodium chloride, calculate the osmotic pressure of sea water. Assume complete ionization of the salt.
Solution. Mass of NaCl = 3.5 g

Actual number of moles of particles of solute in solution=
Volume of solution = 1 litre

Example 33. Molality of a solution in aqueous medium is 0.8. Calculate its mole fraction and the percentage by mass of solute if molar mass of solute is 60.
Solution. We know that,

where, x_B = mole fraction of solute
m_A = molar mass of solvent

x_B = 0.014
Let w_B = x g, w_A = 100 g

Example 34. A very small amount of a non-volatile solute (that does not dissociate) is dissolved in 56.8 cm³ of benzene (density 0.889 g cm^-3). At room temperature, vapour pressure of this solution is 98.8 mm Hg while that of benzene is 100 mm Hg. Find the molality of the solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene, what is the value of molal freezing point depression constant of benzene?
Solution. 

ΔT_f = K_f × Molality
0.73 = K_f × 0.1557
K_f = 4.688

Example 35. x g of a non-electrolytic compound (molar mass = 200) is dissolved in 1.0 litre of 0.05 M NaCl solution. The osmotic pressure of this solution is found to be 4.92 atm at 27ºC. Calculate the value of `x'. Assume complete dissociation of NaCl and ideal behaviour of this solution.
(a) 16.52 gm      
(b) 24.032 gm
(c) 19.959 gm    
(d) 12.35 gm
Ans. (c)
Solution. 
(i) For NaCl : p = iCRT = 2 × 0.05 × 0.0821 × 300 = 2.463 atm
(ii) For unknown compound,

Total osmotic pressure p = p₁ + p₂
4.92 = 2.463 + 0.1231 x
x = 19.959 g

Example 36. The freezing point of a solution containing 50 cm³ of ethylene glycol in 50 g of water is found to be -34ºC. Assuming ideal behaviour, calculate the density of ethylene glycol (K_f for water = 1.86 K kg mol^-1).
(a) 1.13 g/cm³
(b) 2.00 g/cm³
(c) 1.8 g/cm³
(d) 2.25 g/cm³
Ans. (a)
Solution. 

Example 37. Match the boiling point with K_b for x, y and z if molecular weight of x, y and z are same.

Solution. Molal elevation constant may be calculated as,

 
(where, Tº_b = boiling point of pure solvent L_v = latent heat of vaporization.)

(here, ΔH_V = molar latent heat of vaporization).

here, ΔS_V = entropy of vaprization.
By considering ΔS_V as almost constant, K_b µ Tº.
K_b(x) = 0.68 ; K_b (y) = 0.53 and K_b (z) = 0.98.

Example 38. 1.22 g C₆H₅COOH is added into two solvents and data of ΔT_b and K_b are given as -
(a) ln 100 g CH₃COCH₃; ΔT_b= 0.17; K_b = 1.7 kg kelvin/mol
(b) ln 100 g benzene; ΔT_b = 0.13; K_b = 2.6 kg kelvin/mol
Find out the molecular weight of C₆H₅COOH in both cases and inerpret the result.
Solution. 
(a) 

(b) 

(Abnormally double molecular mass of benzoic acid, it shows association of benzoic acid in benzene).

Example 39. How much C₂H₅OH should be added to 1 litre H₂O so that it will not freeze at -20ºC?
K_f = 1.86ºC/m
Solution. Mass of 1 litre water = 1000 g

Example 40. Calculate the molarity of each of the ions in solution when 3.0 litre of 4.0 M NaCl and 4.0 litre of 2.0 M CoCl₂ are mixed and diluted to 10 litre.
Solution. Molarity Na  = 1.2 M
Molarity Co²⁺  = 0.8 M
Molarity Cl^- = 2.8 M

Total Cl^- ions = 28 mole.

Example 41. Calculate the molarity of each ion in solution after 2.0 litre of 3.0 M AgNO₃ is mixed with 3.0 litre of 1.0 M BaCl₂.
Solution. 
BaCl₂₊  2AgNO₃ → 2AgCl +Ba(NO₃)₂
Initial 3 6 -_
mole
Final - -6 (ppt) 3
mole

Example 42. 1.2 kg ethylene glycol  was added in a car radiator containing 9 litre water. The freezing of water was just prevented when car was running in the Himalayan valley at temperature -4ºC. Sudden thunderstorm in the valley lowered the temperature to -6ºC. Calculate the amount of ice separated.
Solution. 

A → Solute; B → Solvent

w_B = 6000 g
weight of ice = (Total weight of H₂O) - (wt. of H₂O at 6°C) = 9000 - 6000 = 3000 g = 3 kg

Example 43. The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is to be 20 mm of mercury.
(a) 0.8
(b) 0.6
(c) 0.4
(d) 0.2
Ans. (b)
Solution. 
Mole fraction of solute 

Comparing under the two conditions,

or mole fraction of solute = 0.4
mole fraction of solvent = (1 -0.4) = 0.6

Example 44. Three solutions of HCl having normality 12 N, 6 N and 2 N are mixed to obtain a solutions of 4 N normality. Which among the following volume ratio is correct for the above three components?
(a) 1 : 1 : 5
(b) 1 : 2 : 6
(c) 2 : 1 : 9
(d) 1 : 2 : 4
Ans. (b)
Solution. 
Use Hit & Trial Method.
N₁V₁ + N₂V₂ + N₃V₃ = N_R(V₁ + V₂ + V₃)
12 × 1 6 × 2 2 × 6 = N_R(9)
N_R = 4

Example 45. Two solutions of H₂SO₄ of molarities x and y are mixed in the ratio of V₁ mL : V₂ mL to form a solution of molarity M₁. If they are mixed in the ratio of V₂mL : V₁mL, they form a solution of molarity M₂. Given V₁. Given V₁/V₂ => 1 and  = , then x : y is -
(a) 2 : 1
(b) 4 : 1
(c) 1 : 2
(d) 3 : 1
Ans. (a)
Solution.
Molarity of the mixture can be calculated as.
M₁V₁ + M₂V₂ = M_R(V₁ + V₂)
where, M_R = resultant solution
(V₁ × x) × (V₂ × y) = M₁(V₁ + V₂)
(V₂ × x) × (V₁ × y) = M₂(V₁ + V₂)
Dividing equation (i) by equation (ii), we get

Substituting we can calculate x : y.

Example 46. Isuling (C₂H₁₀O₅)_n is dissolved in a suitable solvent and the osmotic pressure (p) of solutions of various concentrations (g/cc) C is measured at 20ºc. The slope of the plot of p against 'C' is found to be 4.65 × 10^-3. The molecular weight of insulin is -
(a) 4.8 × 10⁵
(b) 9 × 10⁵
(c) 3 × 10⁵
(d) 5.17 × 10⁶
Ans. (d)
Solution.

where C = concentration in g/cc,
Comparing eqs. (i) and (ii), ...(ii)

Example 47. Compound PdCl₄.6H₂O is a hydrated complex; 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex (K_f for water = 1.86 K kg mol^-1)
(a) [Pd(H₂O)₆]Cl₄
(b) [Pd(H₂O)₄Cl₂]Cl ₂.2H₂O
(c) [Pd(H₂O)₃Cl₃]Cl.3H ₂O
(d) [Pd(H₂O)₂Cl₄].4H ₂O
Ans. (c)
Solution.
ΔT = i × K_f × m
(273 -269.28) = i × 1.86 × 1
3.72 = i × 1.86
i = 2

Thus, the complex should give two ions in the solution, i.e., the complex will be [Pd(H₂O)₃Cl₃]Cl.3H₂O]

Example 48. pH of 0.1 M monobasic acid is measured to be 2. Its osmotic pressure at a given temperature T K is -
(a) 0.1 RT
(b) 0.11 RT
(c) 1.1 RT
(d) 0.01 RT
Ans. (b)
Solution.
HA  H⁺  A^-
t = 0  C  0     0
t_eq C -C∝ C∝ C∝
[H ] = Ca, [H ] = 10^-pH
Cα = 10^-2
0.1 α = 10^-2
a = 0.1
a =  ; 0.1 =
i = 1.1
p = iCRT
= 1.1 × 0.1 × RT = 0.11 RT

Example 49. Lowering of vapour pressure in 1 molal aqueous solution at 100ºC is -
(a) 13.44 mm Hg
(b) 14.12 mm Hg
(c) 31.2 mm Hg
(d) 35.2 mm Hg
Ans. (a)
Solution.
Molality and mole fraction are related as follows:

m_A = molar mass of solvent
x_B = 0.0176, x_A = 0.9824
p = p°_Ax_A
p = 760 × 0.9824 = 746.624
Δp = p°_A - p = 760 - 746.624 = 13.4 mm Hg.

Example 50. Calculate normality of the mixture obtained by mixing 100ml of 0.1N HCl and 50ml of 0.25N NaOH solution.
(a) 0.0467 N 
(b) 0.0367 N 
(c) 0.0267 N 
(d) 0.0167 N
Ans. (d)
Solution.

Molal eq. of HCl = 100 × 0.1 = 10
Molal eq. of NaOH = 50 × 0.25 = 12.5
HCl and NaOH neutralize each other with equal eq.
Eq. of NaOH left = 12.5 - 10 = 2.5
Volume of new solution = 100 + 50 = 150 ml.

Hence normality of the mixture obtained is 0.0167 N 

Example 51. 300 ml 0.1 M HCl and 200 ml of 0.03M H₂SO₄ are mixed. Calculate the normality of the resulting mixture-
(a) 0.084 N 
(b) 0.84 N 
(c) 2.04 N 
(d) 2.84 N 
Ans. (a)
Solution.
For HCl For H₂SO₄
V₁ = 300 ml V₂ = 200 ml
N₁ = M × Basicity N₂ = M × Basicity
= 0.1 × 1 = 0.1 = 0.03 × 2 = 0.06
Normality of the mixture

 = 0.084 N 

Example 52. In what ratio should a 6.5 N HNO₃ be diluted with water to get 3.5 N HNO₃?
(a) 6 : 7 
(b) 7 : 6 
(c) 5 : 6 
(d) 6 : 5
Ans. (b)
Solution.
N₁V₁ = N₂V₂
6.5 V₁ = 3.5 (V₁ x)
6.5 V₁ = 3.5 V₁ 3.5 x
3 V₁ = 3.5 x
 

Example 53. Calculate the amount of each in the following solutions -
(i) 150 ml of N/7 H₂SO₄ 
(ii) 250 ml of 0.2M NaHCO₃
(iii) 400 ml of N/10 Na₂CO₃ 
(iv) 1052 g of 1 m KOH.
(a) 52g, 2.12g, 4.2g, 1.05g 
(b) 1.05g, 4.2g, 2.12g, 52g
(c) 1.05g, 2.12g, 52g, 4.2g 
(d) 4.2g, 2.12g, 1.05g, 52g
Ans. (b)
Solution.
(i) Eq. wt. of H₂SO₄ 

Amount of H₂SO₄ per litre (strength) = Normality × Eq. wt. = × 49 = 7 g/litre
Amount in 150 ml  

(ii) Molecular wt. of
NaHCO₃ = 23 1 12 48 = 84
Amount of NaHCO₃ required to produce 1000 c.c. of one molar solution = 84 g
Amount present per litre in 0.2 M solution = 84 × 0.2 = 16.8 g
Amount present in 250 c.c.

 = 4.2 g
(iii) Equivalent weight of

= 53
Amount of Na₂CO₃ = Normality × Eq. wt. = × 53 = 5.3 g/litre
Amount present in 400 c.c.

= 2.12 g
(iv) We know that 1 molal solution of a substance contains 1000 g of solvent.
Wt. of KOH in 1052 g of 1 m KOH solution = 1052 - 1000 = 52 g

Example 54. How many kilograms of wet NaOH containing 12% water are required to prepare 60 litres of 0.50 N solution ?
(a) 1.36 kg 
(b) 1.50 kg 
(c) 2.40 gm 
(d) 3.16 kg
Ans. (a)
Solution.
One litre of 0.50 N NaOH contains = 0.50 × 40g = 20 g = 0.020 kg
60 litres of 0.50 N NaOH contain
= 0.020 × 60 kg = 1.20 kg NaOH
Since the given NaOH contains 12% water, the amount of pure NaOH in 100 kg of the given NaOH = 100 - 12 = 88 kg
Thus 88 kg of pure NaOH is present in 100 kg wet NaOH
1.20 kg of pure NaOH is present in

= 1.36 kg wet NaOH 

Example 55. Calculate the vapour pressure of a solution at 100⁰C containing 3g of cane sugar in 33g of water. (At wt. C = 12 , H = 1 , O = 16)
(a) 760 mm 
(b) 756.90 mm 
(c) 758.30 mm 
(d) None
Ans. (b)
Solution.
Vapour pressure of pure water (solvent) at 1000C, p° = 760 mm.
Vapour pressure of solution, p = ?
Wt. of solvent, W = 33g
Wt. of solute, w = 3g
Mol. wt. of water (H₂O), M = 18
Mol. wt. of sugar (C₁₂H₂₂O₁₁),
m = (12 × 12) + (22 × 1) + (11 × 16) = 342
According to Raoult's law,


(pº for H₂O = 760 mm)
= 760 - 3.19 = 756.90 mm

Example 56. Osmotic pressure of a sugar solution at 24°C is 2.5 atmospheres. Determine the concentration of the solution in gm mole per litre.
(a) 0.0821 moles/litre 
(b) 1.082 moles/litre
(c) 0.1025 moles/litre 
(d) 0.0827 moles/litre
Ans. (c)
Solution.
Here it is given that
p = 2.5 atm, T = 24 273 = 297K, S = 0.0821 lit. atm. deg^-1 mol^-1, C = ?
We know that p = CST
or

= 0.1025 moles/litre

Example 57. Twenty grams of a substance were dissolved in 500 ml. of water and the osmotic pressure of the solution was found to be 600 mm of mercury at 15ºC. Determine the molecular weight of the substance-
(a) 1120 
(b) 1198 
(c) 1200 
(d) None of these
Ans. (c)
Solution.
Here it is given that
w = 20 gm ; V = 500 ml.
= 500/1000 = 0.5 litre
p = 600 mm = 600/760  atm;
T = 15 273 = 288⁰A
m = ?
According to Van't Hoff equation,
pV = nST pV


= 1198

Example 58. Blood plasma has the following composition (milli-equivalents per litre). Calculate its osmotic pressure at 37⁰C.
Na  = 138 , Ca²  = 5.2, K  = 4.5, Mg²  = 2.0 , Cl¯ = 105, HCO₃¯ = 25, PO₄³-= 2.2 , SO₄²-= 0.5,
Proteins = 16, Others = 1.0
(a) 7.47 atm 
(b) 7.30 atm 
(c) 7.29 atm 
(d) 7.40 atm
Ans. (a)
Solution.
Since for calculating osmotic pressure we require millimoles/litre therefore
Na⁺= 138 Ca²⁺ = 5.2/2 = 2.6, K = 4.5,
Mg²⁺ = 2.0/2 = 1.0 , Cl¯ = 105,
HCO₃¯ = 24,PO₄^3-=  2.2/3 = 0.73,
SO₄^2-= 0.5/2 = 0.25 , Proteins = 16,
others = 1.0
Total = 294.18 millimoles/litre = 294.18/1000
= 0.294 moles/litre
Now since p = CST
= 0.294 × 0.0821 × .310 = 7.47 atm

Example 59. 0.15g of a substance dissolved in 15g of solvent boiled at a temperature higher by 0.216⁰C than that of the pure solvent. Calculate the molecular weight of the substance. Molal elevation constant for the solvent is 2.16⁰C.
(a) 216 
(b) 100 
(c) 178 
(d) None of these
Ans. (b)
Solution.
Here it is given that
w = 0.15 g, DT_b = 0.216⁰C
W = 15g K_b = 2.16⁰C
m = ?
Substituting values in the expression,

= 100

Example 60. A solution of 0.450 gm of urea (mol. wt 60) in 22.5 g of water showed 0.170⁰C of elevation in boiling point. Calculate the molal elevation constant of water-
(a) 0.17ºC 
(b) 0.45ºC 
(c) 0.51ºC 
(d) 0.30ºC 
Ans. (c) 
Solution.Fig: Structure of urea

Wt. of solute, w = 0.450 g
Wt. of solvent, W = 22.5 g
Mol. wt of solute, m = 60
Molal elevation constant K_b = ?
Boiling point elevation, DT_b = 0.170⁰C
Substituting these values in the equation

=
= 0.51⁰C

Example 61. Calculate the boiling point of a solution containing 0.45g of camphor (mol. wt. 152) dissolved in 35.4g of acetone (b.p. 56.3⁰C); K_b per 100 gm of acetone is 17.2⁰C.
(a) 56.446°C 
(b) 52.401°C 
(c) 56.146°C 
(d) 50.464°C
Ans. (a)
Solution.
Here it is given that
w = 0.45 g, W = 35.4, m = 152,
Kb = 17.2 per 100gm
Now we know that ΔT_b

(Note that this is expression when K_b is given per 100g of the solvent)
Substituting the values in the above expression.

= 0.146⁰C
Now we know that
B.P. of solution (T) - B.P. of solvent (T₀) = DT
B.P. of solution (T) = B.P. of solvent(T₀) DT
Hence B.P. of solution = 56.3 0.146
= 56.446⁰C

Example 62. The freezing point of 0.2 molal K₂SO₄ is _1.1ºC. Calculate Van't Haff factor and percentage degree of dissociation of K₂SO₄. K_f for water is 1.86º
(a) 97.5 
(b) 90.75 
(c) 105.5 
(d) 85.75 
Ans. (a)
Solution.
ΔT_f = freezing point of water -freezing point of solution = 0º C - (1.1º C) = 1.1º
We know that,
ΔT_f = i × K_f × m
1.1 = i × 1.86 × 0.2

But we know
i = 1 (n - 1) a
2.95 = 1 (3 - 1) a = 1 2a
a = 0.975
Van't Haff factor (i) = 2.95
Degree of dissociation = 0.975
Percentage degree of dissociation = 97.5