Difference between revisions of "Lab Molarity"

Latest revision as of 13:59, 21 March 2020

Welcome to your "sort of" first online lab. I would like to use this first lab activity to cover some additional lecture material and then follow it up with a short activity in "lecture" on Weds.

Section 4.5: Concentration of Solutions

Molarity (M)

You may have noticed in lab that many of the solutions we used had a label like "3M" to indicate the concentration of the solute in the solution. 3M means 3 molar or the molarity is 3. As you will see in your text the definition of molarity (M) is the moles of solute divided by liters of solution.

Example 1

What is the molarity of a solution made by adding 1.23 moles of sodium chloride to 1 .00 L of water?

Answer: 1.23 M NaCl

Example 2

What is the molarity of a solution made by adding 12.34 grams of sodium chloride to 1.000 L of water? (Note: you will need to calculate the molecular mass of NaCl...)

molecular mass of NaCl...

12.34 g of NaCl = ? moles of NaCl

then calculate M = ?

Answer: 0.2112 M NaCl

Example 3

What is the molarity of a solution made by adding 43.21 grams of sodium chloride to 1,234 mL of water?

Answer: 0.5992 M NaCl

Example 4

Molarity does not just apply to molecular or ionic compounds, but also ions. What is the molarity of Na⁺ and Cl^- ions in a solution made by adding 43.21 grams of sodium chloride to 1,234 mL of water?

Given that you have completed Example 3, we know that the molarity of NaCl = 0.5992 M

Answer: [Na⁺] = [Cl^-]= 0.5992 M NaCl

Notice that we use brackets around the chemical formula, ie. [NaCl] to mean "molarity of NaCl"...this is shorthand notation.

Example 5

What is the molarity of Ca²⁺ and Cl^- ions in a solution made by adding 34.2 grams of calcium chloride to 543 mL of water?

Answer: [Ca⁺] = 0.568 M and [Cl^-]= 1.14 M

The above examples are considered fairly basic. In these problems above you were given either moles or grams of solute and the final volume of the solution and asked to calculate the molarity. There are 3 values, grams/moles (either one), volume of solution, and molarity; as with any math problem, if given any two fo the three, you can solve for the other.

Example 6

How many moles of NaCl are in 1.234 L of a 0.521 M NaCl solution? (...follow the units...)

<write it out>...here is how i did it...

Example 7

How many grams of NaCl are in 1.234 L of a 0.521 M NaCl solution? (...follow the units...)

<write it out>...this is an extension of the problem above...first calculate moles then grans...

Answer = 37.6 grams NaCl

Now go to WebAssign and complete HW4.2 spring 2020 (Molarity) ,<-- not available until Mon at 9 am

END of "not" Lab :)

@@ Line 1: / Line 1: @@
-Welcome to your "sort of" first online lab. I would like to use this first lab activity to cover some additional lecture material and then follow it up with a short activity.
+Welcome to your "sort of" first online lab. I would like to use this first lab activity to cover some additional lecture material and then follow it up with a short activity in "lecture" on Weds.
 ==Section 4.5: Concentration of Solutions==
+===Molarity (M)===
 You may have noticed in lab that many of the solutions we used had a label like "3M" to indicate the concentration of the solute in the solution. 3M means 3 ''molar'' or the ''molarity'' is 3. As you will see in your text the definition of molarity (M) is the ''moles of solute'' divided by ''liters of solution''.
 :[[File:Molarity.jpg|300px]]
-====Easy Example====
+====Example 1====
 :What is the ''molarity'' of a solution made by adding 1.23 moles of sodium chloride to 1 .00 L of water?
 ::::<write this out>
 <div align="right"> Answer: 1.23 M NaCl</div>
-====Medium Example====
+====Example 2====
 :What is the ''molarity'' of a solution made by adding 12.34 grams of sodium chloride to 1.000 L of water? (''Note: you will need to calculate the molecular mass of NaCl...'')
 ::::<write this out>
@@ Line 18: / Line 19: @@
 <div align="right"> Answer: 0.2112 M NaCl</div>
-====Hard (but not really) Example====
+====Example 3====
 :What is the ''molarity'' of a solution made by adding 43.21 grams of sodium chloride to 1,234 mL of water?
 ::::<write this out>
 <div align="right"> Answer: 0.5992 M NaCl</div>
+====Example 4====
+:Molarity does not just apply to molecular or ionic compounds, but also ions. What is the ''molarity'' of Na<sup>+</sup> and Cl<sup>-</sup> ions in a solution made by adding 43.21 grams of sodium chloride to 1,234 mL of water?
+::::Given that you have completed Example 3, we know that the molarity of NaCl = 0.5992 M
+<div align="right"> Answer: [Na<sup>+</sup>] = [Cl<sup>-</sup>]= 0.5992 M NaCl
+''Notice that we use brackets around the chemical formula, ie. [NaCl] to mean "molarity of NaCl"...this is shorthand notation.''
+</div>
+====Example 5====
+:What is the ''molarity'' of Ca<sup>2+</sup> and Cl<sup>-</sup> ions in a solution made by adding 34.2 grams of calcium chloride to 543 mL of water?
+::::<write it out>
+<div align="right"> Answer: [Ca<sup>+</sup>] = 0.568 M and [Cl<sup>-</sup>]= 1.14 M </div>
+The above examples are considered fairly basic. In these problems above you were given either moles or grams of solute and the final volume of the solution and asked to calculate the molarity. There are 3 values, grams/moles (either one), volume of solution, and molarity; as with any math problem, if given any two fo the three, you can solve for the other.
+====Example 6====
+:How many moles of NaCl are in 1.234 L of a 0.521 M NaCl solution? (''...follow the units...'')
+::::<write it out>...here is how i did it...
+<div align="center"> [[File:IMG_0293.jpg|400px]]</div>
+====Example 7====
+:How many grams of NaCl are in 1.234 L of a 0.521 M NaCl solution? (''...follow the units...'')
+::::<write it out>...this is an extension of the problem above...first calculate moles then grans...
+<div align="right"> Answer = 37.6 grams NaCl</div>
+<div align="center"> '''Now go to WebAssign and complete HW4.2 spring 2020 (Molarity) ,<-- not available until Mon at 9 am'''</div>
+END of "not" Lab :)