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At 100 Β° C , Β°C, Β° C , the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K b = 0.52 , Β theΒ boilingΒ pointΒ ofΒ thisΒ solutionΒ willΒ be K_b = 0.52, \text { the boiling point of this solution will be} K b β = 0.52 , Β theΒ boilingΒ pointΒ ofΒ thisΒ solutionΒ willΒ be
hard
Solutions
2016
chemistry
Explanation To solve this problem, we need to determine the boiling point elevation of the solution. \\
Given: \\
β’ Mass of solute = 6.5 = 6.5 = 6.5 g \\
β’ Mass of solvent (water) = 100 = 100 = 100 g \\
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β’ Vapour pressure of solution at
100 β C = 732 100^\circ C = 732 10 0 β C = 732 mm Hg
β’ Vapour pressure of pure water at
100 β C = 760 100^\circ C = 760 10 0 β C = 760 mm Hg
β’ Boiling point elevation constant
K b = 0.52 K_b = 0.52 K b β = 0.52
β C β
^\circ C \cdot β C β
kg/mol
First, calculate the mole fraction of the solute using Raoult's Law:
P 0 β P = P 0 β
X s o l u t e \\
P_0 - P = P_0 \cdot X_{solute} \\
P 0 β β P = P 0 β β
X so l u t e β Where:
P 0 = 760 \\
P_0 = 760 P 0 β = 760 mm Hg (vapour pressure of pure water)
mm Hg (vapour pressure of solution)
Substitute the values:
760 β 732 = 760 β
X s o l u t e 28 = 760 β
X s o l u t e X s o l u t e = 28 760 X s o l u t e β 0.0368 \\
760 - 732 = 760 \cdot X_{solute} \\
28 = 760 \cdot X_{solute} \\
X_{solute} = \frac{28}{760} \\
X_{solute} \approx 0.0368 \\
760 β 732 = 760 β
X so l u t e β 28 = 760 β
X so l u t e β X so l u t e β = 760 28 β X so l u t e β β 0.0368 Next, calculate the molality of the solution:
Mole fraction of solute
X s o l u t e = n s o l u t e n s o l u t e + n s o l v e n t X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \\
X so l u t e β = n so l u t e β + n so l v e n t β n so l u t e β β Assuming the molar mass of the solute is
then:
n s o l u t e = 6.5 M n s o l v e n t = 100 18 \\
n_{solute} = \frac{6.5}{M} \\
n_{solvent} = \frac{100}{18} n so l u t e β = M 6.5 β n so l v e n t β = 18 100 β (moles of water)
Substitute into the mole fraction equation:
0.0368 = 6.5 M 6.5 M + 100 18 \\
0.0368 = \frac{\frac{6.5}{M}}{\frac{6.5}{M} + \frac{100}{18}} \\
0.0368 = M 6.5 β + 18 100 β M 6.5 β β Solve for
M : 0.0368 ( 6.5 M + 100 18 ) = 6.5 M 0.0368 β
6.5 M + 0.0368 β
100 18 = 6.5 M 0.0368 β
6.5 M + 0.2044 = 6.5 M 0.2044 = 6.5 M β 0.0368 β
6.5 M 0.2044 = 6.5 ( 1 β 0.0368 ) M 0.2044 = 6.5 β
0.9632 M M = 6.5 β
0.9632 0.2044 M β 30.6 M: \\
0.0368 \left(\frac{6.5}{M} + \frac{100}{18}\right) = \frac{6.5}{M} \\
0.0368 \cdot \frac{6.5}{M} + 0.0368 \cdot \frac{100}{18} = \frac{6.5}{M} \\
0.0368 \cdot \frac{6.5}{M} + 0.2044 = \frac{6.5}{M} \\
0.2044 = \frac{6.5}{M} - 0.0368 \cdot \frac{6.5}{M} \\
0.2044 = \frac{6.5(1 - 0.0368)}{M} \\
0.2044 = \frac{6.5 \cdot 0.9632}{M} \\
M = \frac{6.5 \cdot 0.9632}{0.2044} \\
M \approx 30.6 M : 0.0368 ( M 6.5 β + 18 100 β ) = M 6.5 β 0.0368 β
M 6.5 β + 0.0368 β
18 100 β = M 6.5 β 0.0368 β
M 6.5 β + 0.2044 = M 6.5 β 0.2044 = M 6.5 β β 0.0368 β
M 6.5 β 0.2044 = M 6.5 ( 1 β 0.0368 ) β 0.2044 = M 6.5 β
0.9632 β M = 0.2044 6.5 β
0.9632 β M β 30.6 g/mol
Now, calculate the molality m
m = n s o l u t e m a s s Β o f Β s o l v e n t Β i n Β k g m = 6.5 / 30.6 0.1 m β 2.12 \newline m = \frac{n_{solute}}{mass \ of \ solvent \ in \ kg} \\
m = \frac{6.5/30.6}{0.1} \\
m \approx 2.12 m = ma ss Β o f Β so l v e n t Β in Β k g n so l u t e β β m = 0.1 6.5/30.6 β m β 2.12 mol/kg
Calculate the boiling point elevation
Ξ T b : Ξ T b = K b β
m Ξ T b = 0.52 β
2.12 Ξ T b β 1.10 β C \Delta T_b: \\
\Delta T_b = K_b \cdot m \\
\Delta T_b = 0.52 \cdot 2.12 \\
\Delta T_b \approx 1.10^\circ C \\
Ξ T b β : Ξ T b β = K b β β
m Ξ T b β = 0.52 β
2.12 Ξ T b β β 1.1 0 β C The boiling point of the solution is:
100 β C + 1.10 β C = 101.10 β C \\
100^\circ C + 1.10^\circ C = 101.10^\circ C \\
10 0 β C + 1.1 0 β C = 101.1 0 β C Since the options are given in whole numbers, the closest option is
102 β C . 102^\circ C. \\
10 2 β C . Therefore, the boiling point of the solution is
102 β C , 102^\circ C, 10 2 β C , which corresponds to Option 1.