1. Module 4 Intro

1.38. Page 4

Lesson 5

Module 4—Properties of Solutions

Read

 

Formulas are often required to be manipulated in order to solve for a desired quantity. The following diagrams show how the two formulas used in these examples can be manipulated.

 

Derivations of the equations for moles and concentration are shown.The equation for moles can be rearranged as: Lower case m equals lower case n multiplied by upper case M and as upper case M equals lower case m divided by lower case n. The equation for concentration can be rearranged as: Lower case n equals lower case c divided by upper case V, and as upper case V equals lower case n divided by lower case c.


 

Use the manipulated formulas as you work through “SAMPLE problem 5.1” and “COMMUNICATION examples” 10 and 11 on pages 209 and 210 of your textbook and Example 3 below.

 

Example 3: What volume is required to make a 0.54-mol/L solution with 21 g of sodium hydroxide?

 

The mass of NaOH is given. Convert the mass to number of moles, and then determine the volume.

 

 

First, determine the number of moles of solute.

 

 

Now, calculate the volume.

 

 

The volume of sodium hydroxide required is 0.97 L.

 

Self-Check

 

SC 4. What mass of sodium sulfate is present in 1.20 L of a 0.23-mol/L solution?

 

SC 5. What mass of magnesium phosphate is present in 5.00 L of a 230-mmol/L solution?

 

SC 6. What volume of solution is required to make a 0.86-mol/L solution with 18 g of ammonium nitrate?

 

SC 7. What volume of solution is required to make a 0.55-mol/L solution with 0.29 kg of sodium hydroxide?

 

Check your work.
Self-Check Answers

 

SC 4.

 

 

First, determine the number of moles of sodium sulfate.

 

 

Now, calculate the mass of sodium sulfate.

 

 

39 g of sodium sulfate is present.

 

SC 5.

 

 

First, determine the number of moles of magnesium phosphate.

 

 

Now, calculate the mass of magnesium phosphate.

 

 

302 g of magnesium phosphate is present.

 

SC 6.

 

 

First, determine the number of moles of solute.

 

 

Now, calculate the volume.

 

 

0.26 L of solution is required.

 

SC 7.

 

 

First, determine the number of moles of solute.

 

 

Now, calculate the volume.

 

 

13 L of solution is required.