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

Chemistry Class 12

Created by: Mohit Rajpoot

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.
All you need of Class 12 at this link: Class 12

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

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

Alternatively, 

Solved Examples - Solutions (Part - 3) Notes | EduRev
(ii) No. of moles of Na2S2O3 
Solved Examples - Solutions (Part - 3) Notes | EduRev
Mass of water = (1250 - 474) = 776 g
No. of moles of water
Solved Examples - Solutions (Part - 3) Notes | EduRev
Mole fraction of Na2S2O3

Solved Examples - Solutions (Part - 3) Notes | EduRev
(iii) No. of moles of Na+  ions
= 2 × No. of moles of Na2S2O3
= 2 × 3 = 6
Molality of Na  ions

Solved Examples - Solutions (Part - 3) Notes | EduRev
No. of moles of S2Solved Examples - Solutions (Part - 3) Notes | EduRev ions = No. of moles of Na2S2O3 = 3
Molality of Solved Examples - Solutions (Part - 3) Notes | EduRevions 

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


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 

Solved Examples - Solutions (Part - 3) Notes | EduRev
Mass per cent of water
= (100 -Mass per cent of glucose)
= (100 -8.59) = 91.41%
(ii) No. of moles of glucose
= Solved Examples - Solutions (Part - 3) Notes | EduRev

Mass of water in kg
= Solved Examples - Solutions (Part - 3) Notes | EduRev

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

(iii) No. of moles of glucose 

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

No. of moles of water 

Solved Examples - Solutions (Part - 3) Notes | EduRev
Mole fraction of glucose

Solved Examples - Solutions (Part - 3) Notes | EduRev
Mole fraction of water

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

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

Solved Examples - Solutions (Part - 3) Notes | EduRev
Now, apply dilution formula

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

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

So,

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

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

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

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


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.

 Solved Examples - Solutions (Part - 3) Notes | EduRev
NNaOH º NHCl

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

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 

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

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

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

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

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

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


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

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

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

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

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

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

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

ΔT = iKf × m

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

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

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


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 Solved Examples - Solutions (Part - 3) Notes | EduRev Na+1  Cl-1
(n = 2)
No. of particles after dissociation = 1 (n -1) a
= 1 (2 -1) × 1 = 2

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

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

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

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

Solved Examples - Solutions (Part - 3) Notes | EduRev
(given, p° = 760 mm, w = 14 g, W = 200 g, M = 18, m = 164)

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

Degree of dissociation 

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

Ca(NO3)2Solved Examples - Solutions (Part - 3) Notes | EduRev Ca2+ + 2NO3-
(n = 3)

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

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

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 

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

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

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