Example 27. The density of a 3M sodium thiosulphate solution (Na_{2}S_{2}O_{3}) 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 S_{2}O_{3}^{2}ions 
Solution.
(i) Mass of 1000 mL of Na_{2}S_{2}O_{3} solution
= 1.25 × 1000 = 1250 g
Mass of Na_{2}S_{2}O_{3} in 1000 mL of 3M solution
= 3 × Mol. mass of Na_{2}S_{2}O_{3}
= 3 × 158 = 474 g
Mass percentage of Na_{2}S_{2}O_{3} in solution
Alternatively,
(ii) No. of moles of Na_{2}S_{2}O_{3}
Mass of water = (1250  474) = 776 g
No. of moles of water
Mole fraction of Na_{2}S_{2}O_{3}
_{}
(iii) No. of moles of Na^{+}^{ } ions
= 2 × No. of moles of Na_{2}S_{2}O_{3}
= 2 × 3 = 6
Molality of Na^{ } ions
No. of moles of S_{2} ions = No. of moles of Na_{2}S_{2}O_{3} = 3
Molality of ions
Example 28. 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 29. The volume of water which must be added to a mixture of 350 cm^{3} 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 30. The mole fraction of CH_{3}OH 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 CH_{3}OH and y mole of water be present in solution.
Mole fraction of CH_{3}OH
So,
Example 31. 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 32. 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 × 10^{23 }= 1.505 × 10^{22}
Example 33. 250 mL of a Na_{2}CO_{3} solution contains 2.65 g of Na_{2}CO_{3}. 10 mL of this solution is added to x mL of water to obtain 0.001 M Na_{2}CO_{3} solution. The value of x is :
(Molecular mass of Na_{2}CO_{3} = 10^{6} amu)
(a) 1000
(b) 990
(c) 9990
(d) 90
Ans. (b)
Solution.
Molarity of solution
M_{1}V_{1} = M_{2}V_{2}
0.1 × 10 = 0.001 (10 x)
x = 990 mL
Example 34. 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.
M_{1}V_{1}_{ } M_{2}V_{2} = M_{R} (V_{1} V_{2})
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 35. 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 36. The mass of a nonvolatile 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 37. 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 38. 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 39. 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
K_{3} [Fe(CN)_{6}]
Example 40. 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 41. van't Hoff factor of Hg_{2}Cl_{2} in its aqueous solution will be (Hg_{2}Cl_{2} is 80% ionized in the solution) 
(a) 1.6
(b) 2.6
(c) 3.6
(d) 4.6
Ans. (b)
Solution.
Example 42. 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 43. The degree of dissociation of Ca(NO_{3})_{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(NO_{3})_{2} Ca^{2+ }+ 2NO_{3}^{}
(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 44. Calculate the normal boiling point of a sample of sea water found to contain 3.5% of NaCl and 0.13% of MgCl_{2} 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 45. 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 46. 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 47. A very small amount of a nonvolatile solute (that does not dissociate) is dissolved in 56.8 cm^{3} 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 48. x g of a nonelectrolytic 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 = p_{1 }+ p_{2}
4.92 = 2.463 + 0.1231 x
x = 19.959 g
Example 49. The freezing point of a solution containing 50 cm^{3} 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/cm^{3}
(b) 2.00 g/cm^{3}
(c) 1.8 g/cm^{3}
(d) 2.25 g/cm^{3}
Ans. (a)
Solution.
Example 50. Match the boiling point with K_{b} for x, y and z if molecular weight of x, y and z are same.
b. pt  Kb  
x  100  0.68 
y  27  0.53 
z  253  0.98 
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 51. 1.22 g C_{6}H_{5}COOH is added into two solvents and data of ΔT_{b} and K_{b} are given as 
(a) ln 100 g CH_{3}COCH_{3}; 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 C_{6}H_{5}COOH 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 52. How much C_{2}H_{5}OH should be added to 1 litre H_{2}O so that it will not freeze at 20ºC?
K_{f} = 1.86ºC/m
Solution. Mass of 1 litre water = 1000 g
Example 53. 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 CoCl_{2} are mixed and diluted to 10 litre.
Solution. Molarity Na^{ } = 1.2 M
Molarity Co^{2+ } = 0.8 M
Molarity Cl^{} = 2.8 M
Total Cl^{} ions = 28 mole.
Example 54. Calculate the molarity of each ion in solution after 2.0 litre of 3.0 M AgNO_{3} is mixed with 3.0 litre of 1.0 M BaCl_{2}.
Solution.
BaCl_{2+ } 2AgNO_{3 }→ 2AgCl +Ba(NO_{3})_{2}
Initial 3 6 _
mole
Final  6 (ppt) 3
mole
Example 55. 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 H_{2}O)  (wt. of H_{2}O at 6°C) = 9000  6000 = 3000 g = 3 kg
Example 56. The vapour pressure of a solvent decreased by 10 mm of mercury when a nonvolatile 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 57. 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.
N_{1}V_{1 }+ N_{2}V_{2 }+ N_{3}V_{3} = N_{R}(V_{1 }+ V_{2 }+ V_{3})
12 × 1 6 × 2 2 × 6 = N_{R}(9)
N_{R} = 4
Example 58. Two solutions of H_{2}SO_{4} of molarities x and y are mixed in the ratio of V_{1} mL : V_{2} mL to form a solution of molarity M_{1}. If they are mixed in the ratio of V_{2}mL : V_{1}mL, they form a solution of molarity M_{2}. Given V_{1}. Given V_{1}/V_{2} => 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.
M_{1}V_{1 }+ M_{2}V_{2} = M_{R}(V_{1 }+ V_{2})
where, M_{R }= resultant solution
(V_{1} × x) × (V_{2} × y) = M_{1}(V_{1 }+ V_{2})
(V_{2} × x) × (V_{1} × y) = M_{2}(V_{1 }+ V_{2})
Dividing equation (i) by equation (ii), we get
Substituting we can calculate x : y.
Example 59. Isuling (C_{2}H_{10}O_{5})_{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 × 10^{5}
(b) 9 × 10^{5}
(c) 3 × 10^{5}
(d) 5.17 × 10^{6}
Ans. (d)
Solution.
where C = concentration in g/cc,
Comparing eqs. (i) and (ii), ...(ii)
Example 60. Compound PdCl_{4}.6H_{2}O 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(H_{2}O)_{6}]Cl_{4}
(b) [Pd(H_{2}O)_{4}Cl_{2}]Cl_{ 2}.2H_{2}O
(c) [Pd(H_{2}O)_{3}Cl_{3}]Cl.3H_{ 2}O
(d) [Pd(H_{2}O)_{2}Cl_{4}].4H_{ 2}O
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(H_{2}O)_{3}Cl_{3}]Cl.3H_{2}O]
Example 61. 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 62. 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°_{A}x_{A}
p = 760 × 0.9824 = 746.624
Δp = p°_{A}  p = 760  746.624 = 13.4 mm Hg.
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