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A glucose solution contains 55.8 g of glucose (C6H12O6) in

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

Solution for problem 105P 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 105P

A glucose solution contains 55.8 g of glucose (C6H12O6) in 455 g of water. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

Step-by-Step Solution:

Solution 105 P:

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

Step 1: Calculation of the amount of glucose in mol

We know,

Molar mass of glucose = 180.15 g/mol

Mass of glucose = 55.8 g

Amount of glucose in mol = 55.8 g/180.1559 g/mol = 0.3097 mol

 Step 2: Calculation of molality of the solution.

Volume of solution = 455 g =0.455 kg

Molality (m) of the solution = 0.3097 mol/0.455 kg = 0.6807 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

        = 0.6807 m x 1.86 oC/m= 1.266 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 - 1.266 oC = -1.266 oC = -1.27oC

Therefore , the freezing point of the solution is  -1.27oC

Step 3 of 3

Chapter 13, Problem 105P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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