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Molarity

The document provides notes on molarity, dilutions, and related calculations in chemistry. It defines molarity as the moles of solute per liter of solution and explains how to calculate it, including examples and practice questions. Additionally, it outlines the dilution equation and provides examples to illustrate the concept of diluting concentrated solutions.

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0% found this document useful (0 votes)
19 views4 pages

Molarity

The document provides notes on molarity, dilutions, and related calculations in chemistry. It defines molarity as the moles of solute per liter of solution and explains how to calculate it, including examples and practice questions. Additionally, it outlines the dilution equation and provides examples to illustrate the concept of diluting concentrated solutions.

Uploaded by

kalsoomume066
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Molarity & Dilutions Notes

Concentration
Measure of how much solute is dissolved in a specific amount of solvent/solution.
 A concentrated solution contains larger amount of solute.
 A dilute solution contains a smaller amount of solute.

Molarity (M)
Moles of solute / liters of solution
 Also known as molar concentration
 It’s unit is M (read/pronounced a molar)

Units Table
Examples:
 What is the molar concentration of a liter solution with 0.5 mol of solute?
Molarity = moles of solute / liter of solution
= 0.5 mol solute / 1 L solution
= 0.5 M
 If you have a 100.5 mL IV (intravenous) solution containing 5.10 g glucose
(C6H12O6). What is the molarity of the solution? (Molar Mass of glucose = 180.16
g/mol)

Molarity = moles of solute / liter of solution


Step 1:
Convert grams of solute glucose to moles solute using MM of glucose
5.10 g glucose x 1 mole glucose = 0.0283 mol glucose
180.16 g glucose
Step 2:
Convert mL of solution to Liters of solution
Step 3:
Solve for molarity
M = moles of solute/ liters of solution
= 0.0283 mol glucose / 0.1005 L solution
= 0.282 M

Dilutions
Adding additional solvent to dilute a more concentrated stock solution.
The total # of moles of solute does not change during a dilution (only solvent is added)
Dilution Equation:

M1V1 = M2V2

M1 = the initial molarity of the concentrated solution


V1 = the initial volume of the concentrated solution
M2 = the final molarity of the dilute solution
V2 = the final molarity of the dilute solution
Example:
What volume of 2.00 M CaCl2 stock solution would you use to make 0.50 L of 0.300 M CaCl2?
M1V1 = M2V2
 M1 = 2.00 M
 V1 = ?
 M2 = 0.300 M
 V2 = 0.50 L
2.00 M x V1 = 0.300M x 0.50L
V1 = 0.075 L

Practice Questions

1) A chemist has 250 mL of a 600 mM hydrochloric acid (HCl) solution. What volume of
water must be added to dilute it to 150 mM?

2) You have 100 mL of a 5.0 M sulfuric acid (H2SO4) solution. How many moles of
H2SO4 are present in this solution?

3) A stock solution of nitric acid (HNO3) has a concentration of 120 mM. How much of this
solution is required to prepare 250 mL of a 20 mM solution?

4) A 25 mL sample of 800 mM sodium hydroxide (NaOH) is diluted to 250 mL. What is the
final concentration of the solution in mM?

5) You mix 150 mL of a 3.0 M potassium chloride (KCl) solution with 100 mL of a 1.5 M
sodium chloride (NaCl) solution. How many moles of chloride ions are present in the
final solution?

6) A laboratory requires 1.2 L of a 200 mM calcium nitrate (Ca(NO3)2) solution. How


many moles of Ca(NO3)2 are needed?

7) A biochemist needs to prepare 750 mL of a 100 mM glucose (C6H12O6) solution from a


1.5 M stock solution. What volume of the stock solution is needed?

8) A 400 mL solution of 750 mM aluminum sulfate (Al2(SO4)3) is diluted to 1.2 L. What is


the new molarity in mM?

9) How much 18 M sulfuric acid (H2SO4) must be used to prepare 500 mL of a 300 mM
solution?

10) A 50 mL solution of 10 M acetic acid (CH3COOH) is diluted to 500 mL. How many
moles of CH3COOH are present in the final solution?

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