Concept explainers
Calculate the vapor pressure of water over each of the following solutions of glycerol, C3H8O3, at 28°C
(vapor pressure of pure water=28.35 mm Hg)
(a)
percent glycerol by mass=30.6%
(b) 2.74 m
(c)
mole fraction of glycerol=0.188
(a)
Interpretation:
To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.
Concept introduction:
Formula to calculate moles of a component is −
Moles=MassMolar mass
Mole fraction of a component in a solution can be found out by dividing moles of the component by summation of mole of all components in the solution.
Mole fraction (X)=Moles of componentSummation of moles of all components
Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.
Psolvent=Xsolvent×P0solvent
Where, Psolvent
Answer to Problem 34QAP
Vapor pressure of water if mass percent of glycerol is 30.6%
Explanation of Solution
Given:
Vapor pressure of pure water = 28.55 mm Hg at 28 °C.
Given mass percent of glycerol is 30.6%
Mass of solvent(water)=100 g−30.6 g=69.4 g
So, mass of water is 69.4 g
Moles of each component of the solution are to be calculated as follows:
Mass of glycerol =30.6 g
Molar mass of glycerol =92.09 g mol−1
Moles of glycerol=Mass of glycerolMolar mass of glycerol=30.6 g92.09 g mol−1=0.33 mol
So, moles of glycerol is 0.33 mol
Mass of water =69.4 g
Molar mass of water =18.02 g mol−1
Moles of water=Mass of waterMolar mass of water=69.4 g18.02 g mol−1=3.85 mol
So, moles of water is 3.85 mol
Calculation of mole fraction of water
Mole fraction of water(Xwater)=Moles of waterMoles of water+Moles of glycerol=3.85 mol3.85 mol+0.33 mol=3.85 mol4.18 mol=0.92
So, mole fraction of water is 0.92
Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.
Psolvent=Xsolvent×P0solvent
Where, Psolvent
Mole fraction of water Xwater=0.92
Vapor pressure of pure solvent =28.35 mm Hg
Vapor pressure of water can be calculated as follows:
Pwater=Xwater×P0water=0.92×28.35 mm Hg=26.08 mm Hg
Hence, vapor pressure of water is 26.08 mm Hg
(b)
Interpretation:
To calculate vapour pressure of water if molality of the solution is 2.74 m
Concept introduction:
Molality is one way to define the concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.
Formula of molality is-
Molality=Moles of soluteMass of solvent(kg)
Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.
Psolvent=Xsolvent×P0solvent
Where, Psolvent
Answer to Problem 34QAP
If molality of the solution is 2.74 m
Explanation of Solution
Given:
Vapor pressure of pure water = 28.55 mm Hg at 28 °C.
Given the molality is 2.74 m
Moles of solute (glycerol) can be calculated as follows:
Molality=Moles of soluteMass of solvent(kg)2.74 m=Moles of solute1 kgMoles of solute=2.74 mol kg−1 kg=2.74 mol
So, moles of solute (glycerol) is 2.74 mol
Moles of solvent can be calculated as follows:
Molar mass of water is =18.02 g mol−1
Mass of solvent (water) =1000 g
Moles of solvent(water)=1000 g18.02 g mol−1=55.5 mol
So, moles of solvent is 55.5 mol
Mole fraction of solvent can be calculated as follows:
Mole fraction of solvent(Xwater)=Moles of solvent (water)Moles of solvent (water)+Moles of solute (glycerol)=55.5 mol55.5 mol+2.74 mol=0.95
So, mole fraction of solvent (water) is 0.95
Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.
Psolvent=Xsolvent×P0solvent
Where, Psolvent
Mole fraction of water Xwater=0.95
Vapor pressure of pure solvent =28.35 mm Hg
Vapor pressure of water can be calculated as follows:
Pwater=Xwater×P0water=0.95×28.35 mm Hg=26.93 mm Hg
Hence, vapor pressure of water is 26.93 mm Hg
(c)
Interpretation:
To calculate vapor pressure of water if mole fraction of glycerol is 0.188
Concept introduction:
For a solution containing solute (glycerol) and solvent (water), summation of mole fraction is equal to one.
Xethylene glycol+Xwater=1
Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.
Psolvent=Xsolvent×P0solvent
Where, Psolvent
Answer to Problem 34QAP
Vapor pressure of water if mole fraction of glycerol is 0.188
Explanation of Solution
Given:
Vapor pressure of pure water = 28.55 mm Hg at 28 °C.
Given mole fraction of glycerol (X glycerol) is 0.188
Addition of mole fraction of all components of a solution is equal to one.
Xethylene glycol+Xwater=1Xwater=1−Xethylene glycol=1−0.188=0.812
So, mole fraction of water is 0.812
Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.
Psolvent=Xsolvent×P0solvent
Where, Psolvent
Mole fraction of water Xwater=0.812
Vapor pressure of pure solvent =28.35 mm Hg
Vapor pressure of water can be calculated as follows:
Pwater=Xwater×P0water=0.812×28.35 mm Hg=23.02 mm Hg
Hence, vapor pressure of water is 23.02 mm Hg
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