IIT PROBLEMS FOR CHAPTER SOLUTIONS LEVEL-1
Example: 50 g of a saturated aqueous solution of potassium chloride at 30°C is evaporated to dryness, when 13.2 g of dry KCl was obtained. Calculate the solubility of KCl in water at 30°C.
Solution: Mass of water in solution = (50 – 13.2) = 36.8 g
Solubility of KCl = Mass of KCl/Mass of water × 100 = 13.2/36.8 × 100
Example: How much copper sulphate will be required to saturate 100 g of a dilute aqueous solution of CuSO4 at 25°C if 10 g of a dilute solution leave on evaporation and drying 1.2g of anhydrous CuSO4? The solubility of CuSO4 in water at 25°C is 25.
Solution: 100 g of dilute solution of CuSO4 contain
= 1.2 × 10 = 12.0 g CuSO4
Mass of water present in dilute solution
= (100 – 12) = 88 g
To saturate 100 g of water, CuSO4 required = 25 g
So, to saturate 88 g of water, CuSO4 required 25/100 × 88
= 22 g
Thus, the mass of CuSO4 to be added to 100 g of dilute solution to saturate it = (22 – 12) = 10 g
Example:
The vapour pressure of pure benzene at a certain temperature is 640mm Hg. A non-volatile solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular mass of the solid substance?
Solution:
According to Raoult’s law.
p0–ps/p0 = n/n+N
Let m be the molecular mass of the solid substance.
n = 2.175/m ; N = 39/78 = 0.5
[molecular mass of benzene = 78]
Substituting the values in above equation.
640–600/640 = 2.175/m/2.175/m+0.5 = 2.174/2.175+0.5m
m = 2.175×16 – 2.175/0.5 = 65.25
Example:
A solution containing 30 g of a non-volatile solute in exactly 90 g of water has a vapour pressure of 21.85 mm of Hg at 25°C. Further 18 g of water is then added to the solution; the new vapour pressure becomes 22.15 mm Hg of at 25°C. Calculate (a) molecular mass o the solute and (b) vapour pressure of water at 25°C.
Solution:
Let the vapour pressure of water at 25°C be p0 and molecular mass of the solute be m.
Using Raoult’s law in the following form.
For solution (I), (p0–21.85)/21.85 = 30×18/90×m …(i)
For solution (II), (p0–22.15)/22.15 = 30×18/108×m …(ii)
Dividing Eq. (i) by Eq. (ii),
(p0–21.85)/21.85 × 22.15/(p0–22.15) = 108/90 = 6/5
Substituting the value of p0 in Eq. (i)
M = 67.9
Example:
What mass of non-volatile solute (urea) needs to be dissolved in 100 g of water in order to decrease the vapour pressure of water by 5%. What will be the molality of solution?
Solution:
Using Raoult’s law in the following form,
p0–ps/ps = wM/Wm
If p0 = 100 mm, then ps = 75 mm
100–75/75 = w×18/100×60
w = 111.1
Molality = w×1000/m×W = 111.1×1000/60×100 = 18.52 m
Example: On dissolving 10.8 glucose (m.w. = 180) in 240 g of water, its boiling point increases by 0.13oC. Calculate the molecular elevation constant of water.
Solution: ?T = 100K'×w/W×m
or K' = ?T×W×m/100×w
Given, ?T =0.13oC, W = 240 g, m = 180 and 2 = 10.8 g
K' = 0.13×240×180/100×10.8 = 5.2o
Example: A solution of 2.5 g of a non-volatile solid in 100 g benzene boiled at 0.42oC higher than the boiling point of pure benzene. Calculate the molecular mass of the substance. Molal elevation constant of benzene is 2.67 K kg mol−1.
Solution: m = 1000Kb×w/W×?T
Given, Kb = 2.67, w = 2.5 g, W = 100 g, ?T = 0.42
Given, Kb = 2.67, w = 2.5 g, W = 100 g, ?T = 0.42
m = 1000×2.67×2.5/100×0.42 = 158.9
The molecular mass of substance is 158.9.
Example: The molal elevation constant for water is 0.56 K kg mol−1. Calculate the boiling point of a solution made bt dissolving 6.0 g of urea (NH2CONH2) in 200 g of water.
Solution: ?T = 1000Kb×w/m×W
Given, Kb = 0.56 K kg mol−1, w = 6.0 g, W = 200 g, m=60
?T = 1000×0.56×6.0/200×60 = 0.28oC
Thus, The boiling point of solution = b.pt. of water + ?T = (100oC + 0.28oC) = 100.28oC
Example: By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7oC. Calculate the molecular mass of the substance. [Molal depression constant for water = 1.863 K kg mol−1]
Solution: m = 1000Kf×w/W×?T
Given, Kf = 1.863 K kg mol−1.
w = 13.6 g, W = 20 g, ?T = 3.7oC
m = 1000×1.863×13.6/20×3.7 = 243.39
No comments:
Post a Comment