diluting solutions. equation m 1 v 1 = m 2 v 2 m = molarity v = volume

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Diluting Solutions

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Example 1 What volume in milliliters of a 2.00M CaCl 2 stock solution would you use to make 0.50L of 0.300M calcium chloride solution? V 1 = ? mL M 1 =2.00M V 2 =0.50L=500m L M 2 =0.300M M 1 V 1 = M 2 V (V 1 ) = 0.300(500) V 1 =75mL **Variables must have the same units!! Look to the word problem for correct number of sig. figs.

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Page 1: Diluting Solutions. Equation M 1 V 1 = M 2 V 2 M = Molarity V = Volume

Diluting Solutions

Page 2: Diluting Solutions. Equation M 1 V 1 = M 2 V 2 M = Molarity V = Volume

Equation

M1 V1 = M2 V2

M = MolarityV = Volume

Page 3: Diluting Solutions. Equation M 1 V 1 = M 2 V 2 M = Molarity V = Volume

Example 1

What volume in milliliters of a 2.00M CaCl2 stock solution would you use to make

0.50L of 0.300M calcium chloride solution? V1= ? mLM1=2.00MV2=0.50L=500mLM2=0.300M

M1 V1 = M2 V2

2.00(V1) = 0.300(500)V1=75mL

**Variables must have the same units!! Look to the word problem for correct number of sig. figs.

Page 4: Diluting Solutions. Equation M 1 V 1 = M 2 V 2 M = Molarity V = Volume

Example 2

What volume in milliliters of a 12.0M HCl solution would you use to make 2.50L of

a 2.00M HCl solution? V1= ? mLM1=12.0MV2=2.50L=2500mLM2=2.00M

M1 V1 = M2 V2

12.00(V1) = 2.00(2500)V1=417mL

Page 5: Diluting Solutions. Equation M 1 V 1 = M 2 V 2 M = Molarity V = Volume

Colligative Properties

The physical properties of solutions that are different than the physical properties of a pure solvent

These depend only upon the number of particles in a solution Vapor pressure lowering Boiling point increasing Freezing point lowering