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Asucrose solution is made using 144 g of sucrose

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

Solution for problem 122P Chapter 13

Introductory Chemistry | 5th Edition

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

Introductory Chemistry | 5th Edition

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Problem 122P

Asucrose solution is made using 144 g of sucrose (C12H22O11) and diluting to a total volume of 1.00 L. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.06 g/mL for the final solution.)

Step-by-Step Solution:

Solution 122P:

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

Step 1: Calculation of the amount of sucrose  in mol

We know,

Molar mass of sucrose  =342.296  g/mol

Mass of sucrose   = 144 g

Amount of glucose in mol = 144g/342.296 g/mol = 0.4207 mol

 Step 2: Calculation of molality of the solution.

Density = Mass/volume = 1.06 g/mL

Volume = 1L= 1000 mL

Mass of solution = Density x volume = 1.06 g/mL x 1000 mL = 1060 g

Volume of solution = 1060g  = 1.06 kg

Molality (m) of the solution = 0.4207 mol/1.06 kg  =0.3968 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

        = 0.3968 m x 1.86 oC/m= 0.7382 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 - 0.7382 oC = -0.7382 oC  

Therefore , the freezing point of the solution is  -0.7382 oC or -0.74 oC

Step 3 of 3

Chapter 13, Problem 122P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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