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Sveta_85 [38]
3 years ago
12

Imagine two solutions with the same concentration and the same boiling point, but one has ethanol as the solvent and the other h

as carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.
Given

Ethanol

normal Boiling point: 78.4

Kb: 1.22

CCL4

normal boiling point: 76.8

Kb: 5.03
Chemistry
1 answer:
defon3 years ago
4 0

Answer: m = 0.42; Tb = 79°

Explanation: The relationship between boiling point of the solvent above a solution is directly proportional to the molal concentration of the solute, i.e.:

ΔT = K_{b}.m

where

ΔT is the change in boiling point of the solvent;

K_{b} is the molal boiling point elevation constant;

m is the molal concentration of the solute in the solution;

For there two solutions:

1) Ethanol:

ΔT = K_{b}.m

Tb - T_{normal} = K_{b}.m

Tb - 78.4 = 1.22.m (1)

2) Carbon Tetrachloride:

Tb - T_{normal} = K_{b}.m

Tb - 76.8 = 5.03.m (2)

Solving the system of equations:

Tb - 78.4 = 1.22.m

Tb = 1.22.m + 78.4 (3)

Substituing (3) in (2)

1.22.m + 78.4 - 76.8 = 5.03m

3.81m = 1.6

m = 0.42

With m, find Tb:

T - 76.8 = 5.03.0.42

T = 2.11 + 76.8

T = 79°

<u>Molal</u> concentration is <u>0.42</u> and boiling point is 79°

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Answer:

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Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

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25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

               

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<h3>Stoichiometry</h3>

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