   Chapter 17, Problem 116IP

Chapter
Section
Textbook Problem

# Some nonelectrolyte solute (molar mass = 142 g/mol) was dissolved in 150. mL of a solvent (density = 0.879 g/cm3). The elevated boiling point of the solution was 355.4 K. What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 kJ/mol, the entropy of vaporization is 95.95 J/K mol, and the boiling-point elevation constant is 2.5 K kg/mol.

Interpretation Introduction

Interpretation: The molar mass of solute, volume, density, enthalpy of vaporization, entropy of vaporization and boiling point elevation constant of solvent is given. The mass of solute that was dissolved in the solvent is to be calculated.

Concept introduction: The expression of elevation in boiling point is,

ΔT=1000×kb×ωm×W

Explanation

Explanation

Given

Enthalpy of vaporization of solvent is 33.90kJ/mol .

Entropy of vaporization of solvent is 95.95J/Kmol

The conversion of kilo-joule (kJ) into joule (J) is done as,

1kJ=103J

Hence,

The conversion of 33.90kJ into joule is,

33.90kJ=(33.90×103)J=33.90×103J

Formula

The entropy of vaporization is calculated as,

ΔS=ΔHvapT

Where,

• ΔHvap is the enthalpy of vaporization.
• ΔS is the entropy of vaporization.
• T is the elevation in boiling point of solvent.

Substitute the values of ΔHvap and ΔS in the above equation.

ΔS=ΔHvapTT=(33.90×10395.95)K=353.30K_

The elevation in boiling point of solvent is 353.3K .

The elevated boiling point of solution is 355.4K .

The expression of boiling point elevation is calculated as,

ΔT=T1T

Where,

• ΔT is the boiling point elevation.
• T1 is the elevated boiling point of solution.

Substitute the values of T1 and T in the above equation.

ΔT=T1T=355.4K353

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 