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Water is the only substance found naturally in all three states on Earth's surface.
At 100°C,the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb=0.52, the boiling point of this solution will be
hard
Solutions
2016
chemistry
Explanation
To solve this problem, we need to determine the boiling point elevation of the solution.Given:• Mass of solute =6.5 g• Mass of solvent (water) =100 g
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Sign up / Login • Vapour pressure of solution at
100∘C=732 mm Hg
• Vapour pressure of pure water at
100∘C=760 mm Hg
• Boiling point elevation constant
Kb=0.52
kg/mol
First, calculate the mole fraction of the solute using Raoult's Law:
P0−P=P0⋅Xsolute Where:
P0=760 mm Hg (vapour pressure of pure water)
mm Hg (vapour pressure of solution)
Substitute the values:
760−732=760⋅Xsolute28=760⋅XsoluteXsolute=76028Xsolute≈0.0368 Next, calculate the molality of the solution:
Mole fraction of solute
Xsolute=nsolute+nsolventnsolute Assuming the molar mass of the solute is
then:
nsolute=M6.5nsolvent=18100 (moles of water)
Substitute into the mole fraction equation:
0.0368=M6.5+18100M6.5 Solve for
M:0.0368(M6.5+18100)=M6.50.0368⋅M6.5+0.0368⋅18100=M6.50.0368⋅M6.5+0.2044=M6.50.2044=M6.5−0.0368⋅M6.50.2044=M6.5(1−0.0368)0.2044=M6.5⋅0.9632M=0.20446.5⋅0.9632M≈30.6 g/mol
Now, calculate the molality m
m=mass of solvent in kgnsolutem=0.16.5/30.6m≈2.12 mol/kg
Calculate the boiling point elevation
ΔTb:ΔTb=Kb⋅mΔTb=0.52⋅2.12ΔTb≈1.10∘C The boiling point of the solution is:
100∘C+1.10∘C=101.10∘C Since the options are given in whole numbers, the closest option is
102∘C. Therefore, the boiling point of the solution is
102∘C, which corresponds to Option 1.