   # A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. The boiling point of this solution was determined to be 77 .85°C. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is 5.03°C ∙ kg/mol, and the boiling point of pure carbon tetrachloride is 76.50°C. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 10, Problem 66E
Textbook Problem
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## A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. The boiling point of this solution was determined to be 77 .85°C. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is 5.03°C ∙ kg/mol, and the boiling point of pure carbon tetrachloride is 76.50°C.

Interpretation Introduction

Interpretation: The molar mass of the given bio-molecule has to be calculated.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution. These properties have a direct relationship to the solute particles, and therefore the colligative properties are useful for identifying the nature of solute particles and also calculating the molar masses of substances.

The change in boiling-point expressed by the given equation

ΔTb=Kbm

Here Kb is molal boiling-point elevation constant.

ΔT is change boiling-point elevation, m is a molality

Molar mass is expressed by

molar mass=massofsubstanceamountofsubstance

### Explanation of Solution

Record the given data from question

The boiling-point constant= 5.03°Ckg/mol

To find the molality of bio-molecule.

The change in boiling-point

ΔTb=77.85°C-76.50°C=1.35°C

Molality of bio-molecule

m=ΔTbKb=1.35°C5.03°Ckg/mol=0.268mol/kg

To find the moles of bio-molecule.

=0.015kgsolvent×0

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