A chemist prepares a solution of aluminum sulfite $(\text{Al}_2(\text{SO}_3)_3)$ by measuring out 179. $\mu\text{mol}$ of aluminum sulfite into a 300. mL volumetric flask and filling the flask to the mark with water.Calculate the concentration in $\text{mmol/L}$ of the chemist's aluminum sulfite solution. Be sure your answer has the correct number of significant digits.A box labeled with the unit $\frac{\text{mmol}}{L}$ is provided. A calculator interface is shown with options for clear, reset, and help.

Molarity It is defined as the number of moles of a substance dissolved in 1 litre of a solution.…

Answered: alculate the concentration in mmol/L of…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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A chemist prepares a solution of aluminum sulfite $(\text{Al}_2(\text{SO}_3)_3)$ by measuring out 179. $\mu\text{mol}$ of aluminum sulfite into a 300. mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in $\text{mmol/L}$ of the chemist's aluminum sulfite solution. Be sure your answer has the correct number of significant digits.

A box labeled with the unit $\frac{\text{mmol}}{L}$ is provided. A calculator interface is shown with options for clear, reset, and help.

Expert Solution

Step 1Introduction

Molarity

It is defined as the number of moles of a substance dissolved in 1 litre of a solution.
It is used to describe the concentration of a solution.

Moles:

$N u m b e r o f m o l e s = W e i g h t i n g r a m s / M o l e c u l a r W e i g h t$

Step 2 Interpretation of information given in the question

According to the question

A chemist prepares a solution of Aluminum sulfite solution by dissolving 179 µmoles of Aluminum sulfite in 300 ml of water.
179 µmoles is equal to 0.000179 moles or 0.179 mmoles
$W e i g h t i n g r a m s = N u m b e r o f m o l e s * M o l e c u l a r W e i g h t = 0.000179 * 342.15 = 0.0612 g m s / 300 m l o f w a t e r$
It means the chemist has dissolved 0.0612 grams of Aluminum sulfite in 300 ml of water.

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