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# A 135-mL sample of a 10.0 M ethylene glycol (C2H6O2) ISBN: 9780321910295 34

## Solution for problem 124P Chapter 13

Introductory Chemistry | 5th Edition

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Problem 124P

A 135-mL sample of a 10.0 M ethylene glycol (C2H6O2) solution is diluted to 1.50 L. What are the freezing and boiling points of the final solution? (Assume a density of 1.05 g/mL for the final solution.)

Step-by-Step Solution:

Solution 124 P:

Here, we are going to calculate the boiling point and freezing point of the final solution .

Step 1: Calculation of the strength of the final solution by using dilution equation.

M1V1= M2V2  ------(1)

Where, M1 and V1 are Molarity and volume of the initial solution whereas M2 and V2 are the Molarity and volume of the solution.

From the Question,

It is given that,

M1 (initial Molarity) =10 M

V1 (initial volume) = 135.0 mL

M2 (final Molarity) =?

V2 (initial volume) = 1.50L =1500 mL

Thus,  from equation (1)

M1V1= M2V2  ------(1) M2 = M1V1/V2

= = 0.9 M

The final strength of  the solution is 0.9 M = 0.9 mol/L

1 L solution contains = 0.9 mol

1.50 L solution contains = 0.9 mol x 1.50L/1 L = 1.35 mol ethylene glycol

Step 2: Calculation of the amount of ethylene glycol  in mol

We know,

The final strength of  the solution is 0.9 M = 0.9 mol/L

1 L solution contains = 0.9 mol

1.50 L solution contains = 0.9 mol x 1.50L/1 L = 1.35 mol ethylene glycol

Step 2: Calculation of molality of the solution.

Density = Mass/volume = 1.05 g/mL

Volume = 1.50L= 1500 mL

Mass of solution = Density x volume = 1.05 g/mL x 1500 mL = 1575 g

Volume of solution = 1575 g    = 1.575 kg

Molality (m) of the solution = 1.35 mol/1.575 kg  =0.8571 m

Step 3: Calculation of freezing point of the solution.

The freezing point of depression of a solution is calculated by the equation: Tf= m x Kf

Where m is the molality of the solution  and Kf is the  freezing point of depression constant for the solvent. Tf= m x Kf , where kf = 1.86 oC/m

=0.8571 m x 1.86 oC/m= 1.594 oC

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 - 1.594 oC = -1.59 oC

Therefore , the freezing point of the solution is  -1.59 oC

Step 3 of 3

##### ISBN: 9780321910295

This full solution covers the following key subjects: solution, final, Density, diluted, Ethylene. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Since the solution to 124P from 13 chapter was answered, more than 353 students have viewed the full step-by-step answer. The answer to “A 135-mL sample of a 10.0 M ethylene glycol (C2H6O2) solution is diluted to 1.50 L. What are the freezing and boiling points of the final solution? (Assume a density of 1.05 g/mL for the final solution.)” is broken down into a number of easy to follow steps, and 37 words. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The full step-by-step solution to problem: 124P from chapter: 13 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM.

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