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An ethylene glycol solution is made using 58.5 g of

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 121P Chapter 13

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

4 5 1 304 Reviews
11
1
Problem 121P

An ethylene glycol solution is made using 58.5 g of ethylene glycol (C2H6O2) and diluting to a total volume of 500.0 mL. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.09 g/mL for the solution.)

Step-by-Step Solution:
Step 2 of 3

Step-1

Given:

Given mass of ethylene glycol (C2H6O2)= 58.5 g

Volume = 500.0 mL

Density = 1.09 g/mL

Know:

Here we have to calculate the the freezing point and boiling point of the solution.

Explanation:

 Moles of C2H6O2 can be calculated as :

(58.5 grams C2H6O2) ( 1 mole C2H6O2 / 62.07 grams) = 0.94 moles of C2H6O2

The  mass of water used can be calculated as,

(500.0 ml) (1.09 grams / ml) = 545 grams of mixture

(545 grams of mixture) - (58.5g of ethylene glycol) = 486.5 grams 0.4865kg of water used

Now  molality is,

(0.94 moles of C2H6O2) / (0.4865 kilograms of water) = 1.932 molal

Step-2

Calculation of freezing point of the solution.

Freezing point ,  Tf= m x Kf

where kf = 1.86 oC/m

 Tf =  (  1.86 C / molal) (1.932)

       =  3.59 C  3.6oC

The freezing-point depression can be calculated as

ΔTf=T∘f−Tf ,

 where T∘f - the freezing point of the pure solvent (water) = 0 oC

Tf - the freezing point of the solution.

 Thus,

Tf  = T∘f - ΔTf = 0 - 3.6 oC = -3.6 oC  

Thus the freezing point will be -3.6oC.

Step-3

Step 3 of 3

Chapter 13, Problem 121P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

The full step-by-step solution to problem: 121P from chapter: 13 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Since the solution to 121P from 13 chapter was answered, more than 638 students have viewed the full step-by-step answer. The answer to “An ethylene glycol solution is made using 58.5 g of ethylene glycol (C2H6O2) and diluting to a total volume of 500.0 mL. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.09 g/mL for the solution.)” is broken down into a number of easy to follow steps, and 41 words. This full solution covers the following key subjects: solution, point, Ethylene, glycol, freezing. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295.

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