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A 250.0-mL sample of a 5.00 M glucose (C6H12O6) solution

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

Problem 123P Chapter 13

Introductory Chemistry | 5th Edition

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

Introductory Chemistry | 5th Edition

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Problem 123P

A 250.0-mL sample of a 5.00 M glucose (C6H12O6) solution is diluted to 1.40 L. What are the freezing and boiling points of the final solution? (Assume a density of 1.06 g/mL for the final solution.)

Step-by-Step Solution:

Problem 123P

A 250.0-mL sample of a 5.00 M glucose (C6H12O6) solution is diluted to 1.40 L. What are the freezing and boiling points of the final solution? (Assume a density of 1.06 g/mL for the final solution.)

Solution 123P:

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) =5 M

V1 (initial volume) = 250.0 mL

M2 (final Molarity) =?

V2 (initial volume) = 1.40L =1400 mL

Thus,  from equation (1)

M1V1= M2V2  ------(1)

                                M2 = M1V1/V2

        

                                 = = 0.8928 M

The final strength of  the solution is 0.8928 M = 0.8928 mol/L

1 L solution contains = 0.8928 mol

1.40 L solution contains = 0.8928 mol x 1.40L/1 L = 1.25 mol sucrose

Step 2: Calculation of the amount of glucose  in mol

We know,

Molar mass of sucrose  =180.156  g/mol

The final strength of  the solution is 0.8928 M = 0.8928 mol/L

1 L solution contains = 0.8928 mol

1.40 L solution contains = 0.8928 mol x 1.40L/1 L = 1.25 mol sucrose

 Step 2: Calculation of molality of the solution.

Density = Mass/volume = 1.06 g/mL

Volume = 1.40L= 1400 mL

Mass of solution = Density x volume = 1.06 g/mL x 1400 mL = 1484 g

Volume of solution = 1484 g   = 1.48 kg

Molality (m) of the solution = 1.25 mol/1.48 kg  =0.84459 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

        = 0.84459 m x 1.86 oC/m= 1.570 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 - 1.570 oC = -1.57 oC

Therefore , the freezing point of the solution is  -1.57 oC

Step 3 of 3

Chapter 13, Problem 123P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo Tro
ISBN: 9780321910295

The full step-by-step solution to problem: 123P from chapter: 13 was answered by Patricia, our top Chemistry solution expert on 05/06/17, 06:45PM. This full solution covers the following key subjects: solution, final, freezing, diluted, boiling. 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 123P from 13 chapter was answered, more than 276 students have viewed the full step-by-step answer. Introductory Chemistry was written by Patricia and is associated to the ISBN: 9780321910295. The answer to “A 250.0-mL sample of a 5.00 M glucose (C6H12O6) solution is diluted to 1.40 L. What are the freezing and boiling points of the final solution? (Assume a density of 1.06 g/mL for the final solution.)” is broken down into a number of easy to follow steps, and 36 words.

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