Question
Asked Feb 26, 2020
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Imagine two solutions with the same concentration and the
Normal boiling point
same boiling point, but one has benzene as the solvent and
Solvent
к, (С/m)
(°C)
the other has carbon tetrachloride as the solvent. Determine
benzene
80.1
2.53
the molal concentration, m (or b), and boiling point, T,.
carbon tetrachloride
76.8
5.03
т
Ть -
||
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Imagine two solutions with the same concentration and the Normal boiling point same boiling point, but one has benzene as the solvent and Solvent к, (С/m) (°C) the other has carbon tetrachloride as the solvent. Determine benzene 80.1 2.53 the molal concentration, m (or b), and boiling point, T,. carbon tetrachloride 76.8 5.03 т Ть - ||

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Expert Answer

Step 1

Colligative properties are the properties which depend on the number of atoms or molecules or ions or particles present in the solution. Some common colligative properties are depression in freezing point, elevation in boiling point, osmotic pressure and relative lowering in vapor pressure.

Elevation in boiling point is a colligative property that can be calculated with the help of concentration of solution. The relation between Elevation in boiling point and concentration of solution can be written as:

Elevation in boiling point = ∆Tb = Tsolute + Tsolvent =  kb x molality 

Step 2

Assume two solutions A and B:

Solution A with benzene:

Boiling point of benzene = 80.1 ◦C

Kb for benzene solvent = 2.53 ◦C x kg / moles

Boiling point of A = TA

For the solution of A:

Chemistry homework question answer, step 2, image 1

Step 3

For the solution of B:

Chemistry homework question answer, step 3, image 1

Since the molality of both solutions is same, hence compare equation (1) and (2) :

Chemistry homework question answer, step 3, image 2

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