Solved Examples on Colligative Properties
1. Find the relative lowering of vapour pressure, if 18 g of glucose is dissolved in 90 g of water.
Solution:
Given,
- Mass of glucose = 18 g
- Mass of water = 90 g
Molar Mass of Glucose = 180 g/mol
Number of moles of glucose(nB) = 18/180 = 0.1
Molar Mass of Glucose = 18 g/mol
Number of moles of water(nA) = 90/18 = 5
Relative lowering ΔP/PA° is equal to XB
XB = nB/(nA + nB)
XB = 0.1/0.1 + 5 = 1/51
Relative Lowering of Vapour Pressure(ΔP/PA°) = 1/51
2. Boiling point elevation of a solution containing sucrose and water is 0.256 °C. The molal elevation constant of water is 0.512 °C/m and molar mass of sucrose is 342 g/mol. What is molality?
Solution:
Given:
- Boiling point elevation (ΔTb) = 0.256 °C
- Molal elevation constant (Kb) = 0.512
(ΔTb) = Kb × m
0.256 = 0.512 × m
m = 0.5 mole/kg
3. Calculate the freezing point depression and the freezing point after adding 100.0 g of table salt to 400.0 g of water. (Kf of water = 1.86)
Solution:
Moles of NaCl = mass/molar mass
Moles of NaCl = 100.0/58.443 = 1.71107 mol
Mass of water = 400.0 g = 0.400 kg
Molalilty(m) = (moles of NaCl)/(mass of water in kg)
m = 1.71107/0.400 = 4.2777
NaCl ⇢ Na+ + Cl–
Van’t Hoff factor(i) = Numberof mole after dissociation/number of mole before dissociation
i = 2
Freezing point depression constant for water Kf = 1.86
Freezing point depression = i × Kf × m = 2 × 1.86 × 4.2777 = 15.9 °C
Freezing point of solution = (freezing point of water – freezing point depression) = 0.0 – 15.9
Freezing point of solution = -15.9 °C
4. If 6.8% w/v of cane sugar is isotonic with 1.52% w/v with Thiocarbamide if the molecular weight of cane sugar is 342 find the molecular weight of Thiocarbamide?
Solution:
In Isotonic solution,
π1 = π2 = i2C2RT
Now, i1C1 = i2C2
For canesugar and thiocarbamide are non electrolytes,
So i = 1
Thus, C1 = C2
% (W/V) percent is the number of grams of solute in 100 mL of solution
C = (Number of moles of solute)/(Volume of solution in L)
C1 = 6.8 × 1000/342 × 100
C2 = 1.52 × 1000/x × 100
As C1 = C2
6.8 × 1000/342 × 100 = 1.52 × 1000/x × 100
x = 76
Thus, weight of Thiocarbamide is 76 g
5. Osmotic pressure, of a solution of glucose, is 117.4 atm. Find the molarity of the solution at 298 K.
Solution:
π = iCRT
(Glucose is a Non Electrolyte, i = 1)
117.4 = C × 0.0821 × 298
C = 4.8 mole/l
Thus, the concentration of glucose is 4.8 moles/litre
6. x grams of solute is dissolved in 500 gram solvent if the sum of elevation of boiling point and depression in freezing point for sucrose in water is 5 find its molality. (if Kb = 0.52 and Kf = 1.86) find x?
Solution:
ΔTf + ΔTb = 5
We now that,
- ΔTf = i × kf × m
- ΔTb = i × kb × m
Here, i = 1 (for nonelectrolyte like sucrose)
kf × m + kb × m = 5
2.38 × m = 5
m = 2.1 mole/kg
Molality = Number of Moles of Solute/Weight of Solvent in Kg
(x × 1000)/(342 × 500) = 2.1
x = 359 gram
Thus, the solute dissolved is 359 grams.
Colligative Properties
Colligative Properties of any solution is the property of the solution that depends on the ratio of the total number of solute particles and the total number of solvent particles. Changing the moles or number of particles of solute or solvent changes the colligative properties of the solution. These colligative properties are not dependent on the chemical nature of the solute or solvent but are rather dependent on the number of solutes and solvent particles in the solution. These colligative properties depend on the number of particles in the solution rather than the nature of the solute and the solvent. These properties can be easily linked with the concentration of the solution, i.e. Molarity, Normality, and Molality.
In this article, we will learn about, various types of Colligative properties of the solution, their examples, and others in detail.
Table of Content
- What are Colligative Properties?
- Colligative Properties Examples
- Types of Colligative Properties
- Lowering Of Vapour Pressure
- Elevation in Boiling Point (ΔTb)
- Depression in Freezing Point (ΔTf)
- Osmotic Pressure (π)
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