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Calculate the molarity of each solution.(a) 33.2 g of KCl

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

Solution for problem 62P 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 62P

Calculate the molarity of each solution.

(a) 33.2 g of KCl in 0.895 L of solution

(b) 61.3 g of C2H6O in 3.4 L of solution

(c) 38.2 mg of KI in 112 mL of solution

Step-by-Step Solution:

Solution 62P :

Step 1:

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

Molarity(M) is a unit of concentration measuring the number of moles of a solute per liter of solution. The SI unit of molarity is mol/L.

        

Formula to find the molarity of solution :

                Molarity =

(a) 33.2 g of KCl in 0.895 L of solution :

Given:

Amount of solute -KCl = 33.2 g

Volume of solution = 0.895 L

We have the amount of solute in g, let’s calculate the number of moles first :

        Number of moles (n) =

Molar mass of KCl = 74.55 g/mol.

Therefore, n =

                = 0.445 moles.

Now, let’s calculate the molarity :

                Molarity =

                Molarity = 0.497 mol/L

Step 2:

(b) 61.3 g of C2H6O in 3.4 L of solution :

Given :

Amount of solute -C2H6O = 61.3 g

Volume of solution = 3.4 L

We have the amount of solute in g, let’s calculate the number of moles first :

        Number of moles (n) =

Molar mass of C2H6O = 46.06 g/mol.

Therefore, n =

                = 1.330 moles.

Now, let’s calculate the molarity :

                Molarity =

                Molarity = 0.391 mol/L

Step 3 of 3

Chapter 13, Problem 62P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Calculate the molarity of each solution.(a) 33.2 g of KCl

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