Examples on Solubility
Example 1: At 313 K, benzene and toluene form perfect solutions A and B. 4 moles of toluene and 1 mole of C6H6 makeup Solution A. Toluene and benzene are equal amounts in Solution B. In each scenario, calculate the total pressure. At 313 K, C6H6 and toluene have vapor pressures of 160 and 60 mm, respectively.
Solution:
- For Solution A
PM = P‘B + P‘T = (P0B × XB) + (P0T × XT)
PM = 160 × (1/1+4) + 60 × (4/1+4)
PM = 32 + 48
PM = 80 mm
- For Solution B
PM = 160 × (92/170) + 60 × (78/170)
PM = 86.588 + 27.529
PM = 114.117 mm
Example 2: Heptane and octane form an ideal solution at 373 K, the vapor pressures of the pure liquids at this temperature are 105.2 kPa and 46.8 kPa respectively. If the solution contains 25g of heptane and 28.5g of octane, calculate the vapor pressure exerted by heptane.
Solution:
Given,
- Po(C7H16) = 105.2 kPa
- Po(C8H18) = 46.8 kPa
- M(C7H16) = 100g mol-1
- M(C8H18) = 114g mol-1
X{C7H16} = n(C7H16) / {n(C7H16) + n(C8H18)}
X{C7H16} = (25/100) / ((25/100) + (28.5/114))
X{C7H16} = 0.25/0.25 + 0.25
X{C7H16} = 0.5
X{C8H18} = 1 – 0.5 = 0.5
P{C7H16} = 105.2 × 0.5 = 52.60 kPa
Solubility
Solubility is a fundamental concept in chemistry that describes the ability of a substance to dissolve in a particular solvent under specific conditions to form a solution. A fluid may or may not dissolve completely in a fluid. Understanding the concept of solubility is essential in many fields of science, including pharmaceuticals, environmental science, and materials science.
In this article, we will explore the key concepts of solubility, such as factors that affect solubility, solubility product, and solubility of different phases of matter with each other.
Table of Content
- What is Solubility?
- Solubility of Liquids In Liquids
- Solubility of Solids In Liquids
- Solubility of Gases In Liquids
- Henry’s Law
- Raoult’s Law
Contact Us