Answered: Calculate the molality, molarity, and…

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter4: Chemical Reactions In Solution

Section: Chapter Questions

Problem 4.35QE: Calculate the molarity of AgNO3 in a solution prepared by dissolving 1.44 g AgNO3 in enough water to...

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Question

Calculate the molality, molarity, and mole fraction of FeCl₃ in a 28.8 mass % aqueous solution (d=1.280 g/mL).

Expert Solution

Step 1

Molarity is equal to the number of moles of solute dissolved per liter of solution. It is denoted by M. Its SI unit is mol/L. Molality is equal to the number of moles of solute dissolved per kg of solvent. It is denoted by m. Its SI unit is mol/kg. Mole fraction is equal to the number of moles of the given substance to the total moles of the substance. It is unitless. The formula for calculating molarity, morality, and mole fraction is shown below.

$\begin{array}{rcl} Molarity & = & \frac{Mass}{Molar mass} \times \frac{1000}{Volume} . \\ Molality & = & \frac{Mass}{Molar mass} \times \frac{1000}{Mass of solvent} . \\ Mole fraction & = & \frac{Number of moles of the given substance}{Total number of moles of the substance} . \end{array}$

Step 2

a)To determine the molarity.

Data given

Mass % of soluion=28.8 %.

Density=1.280 g/mL.

Mass of FeCl₃=28.8 g.

Mass of solvent=71.2 g.

Mass of solution=100 g.

The molar mass of FeCl₃=162.2 g/mol.

To determine the molarity of the solution.

$\begin{array}{rcl} Density & = & \frac{Mass}{Volume} . \\ 1.280 g / mL & = & \frac{100 g}{Volume} . \\ Volume & = & \frac{100 g}{1.280 g / mL} . \\ Volume & = & 78.125 mL . \end{array}$

Molarity of the solution.

$Molarity = \frac{Mass}{Molar mass} \times \frac{1000}{Volume of solution} . Molarity = \frac{28.8 g}{162.2 g / mol} \times \frac{1000}{78.125 mL} . Molarity = 2.27 M .$

Step 3

b)To determine the molality.

$Molality = \frac{Mass}{Molality} \times \frac{1000}{Mass of solvent} . Molality = \frac{28.8 g}{162.2 g / mol} \times \frac{1000}{71.2 g} . Molality = 2.49 m .$

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