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Calculate the molality of a solution containing 12.5 g of

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

Solution for problem 99P 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 99P

Calculate the molality of a solution containing 12.5 g of ethylene glycol (C2H6O2) dissolved in 135 g of water.

Step-by-Step Solution:

Solution 99P :

Step 1:

Here, we have to calculate the molality of each solution :

First, let’s see what is molality :

Molality(m) is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. The SI unit for molality is mol/kg.

Formula to find the molality of solution :

                Molality(m) =

Step 2:

Given :

        Mass of solute - ethylene glycol (C2H6O2)  = 12.5 g

        Mass of solvent - water (H2O) = 135 g

We have solute in grams, we need moles of solute to calculate the molality. Hence, first we find the moles of solute from the grams given :

                Number of moles(n) =

Molar mass of C2H6O2 = 62.07 g/mol

        Therefore, n =

                      n = 0.201 moles

We need mass of solvent in Kg, to calculate molality, hence let’s convert mass of water in g to Kg :

                1 kg = 1000 g

Therefore, 135 g of water in Kg will be :

                = 135g

                = 0.135 Kg.

Step 3 of 3

Chapter 13, Problem 99P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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