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# Solved Examples - Solutions (Part - 3) Notes | EduRev

## Class 12 : Solved Examples - Solutions (Part - 3) Notes | EduRev

The document Solved Examples - Solutions (Part - 3) Notes | EduRev is a part of the Class 12 Course Chemistry Class 12.
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Example 27. 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 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 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 30. 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 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. NNaOH º NHCl 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 × 1023 = 1.505 × 1022

Example 33. 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 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 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 = MR (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 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. xA = 1 -0.0176 = 0.9824
p = p°xA
= 760 × 0.9824 = 746.62
Dp = p° - p = 760 -746.62
= 13.4

Example 36. 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° xA
80 = 100 × xA
xA = 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 Kf and Kfor water are 1.87 kg mol-1 and 0.52 K kg mol-1 respectively -
Solution.
ΔTb = (100.15 -100) = 0.15ºC
We know that, ΔTb = molality × Kb ΔT= molality × Kf
= 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,
Tf = 16.6ºC = 273 16.6 = 289.6 K,
Lf = 180.75 J g-1
Substituting the values in the above equation, Example 39. The freezing point depression of 0.001 m Kx[Fe(CN)6 ] is 7.10 × 10-3 K. Determine the value of x. Given, Kf = 1.86 K kg mol-1 for water -
Solution. Dx = i × Kf × m
7.10 × 10-3 = i × 1.86 × 0.001
i = 3.817 Molecular formula of the compound is
K3 [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 = iKf × m Example 41. 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 42. Calculate the amount of NaCl which must be added to 100 g water so that freezing point is depressed by 2K. For water, Kf = 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(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. Dptheo. = 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, ΔTobs. = 2.4 × ΔTtheo. = 2.4 × 5.84
= 14.02 mm
p -ps = Dpobs. = 14.02
ps = ps -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 MgCl2 by mass. The normal boiling point of water is 100ºC and Kb (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 ΔTb = Molality × Kb
= 1.2846 × 0.51 = 0.655 K
Hence, boiling point of sea water = 373.655 K or 100.655ºC.

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