Molarity By Dilution



p. 31 Molarity By Dilution (mixing a concentrated solvent with water)

Acids are usually acquired from chemical supply houses in concentrated form. These acids are diluted to the desired concentration by adding water. Since moles of acid before dilution = moles of acid after dilution, and moles of acid : CV then,

C1 X V1 = C2 X V2. Solve the following problems.

1) 83 mL 2) 17 mL 3) 130 mL 4)1.0 X102 mL 5) 1100 mL

|1) C1 |

|How much concentrated 18 M sulphuric acid is needed to prepare |

|250.0 mL of a 6.0 M solution? |

|V2 C2 |

|C1V1 =C2V2 ( V1 = C2V2 |

|C1 |

|C1=18 M –concentrated |

|V1= how much volume? |

|C2=6.0 M -diluted |

|V2=250.0 mL –total volume |

| |

|V1 = 6.0 M X 250.0 mL = 83 mL or 0.083 L |

|18 M |

|2) How much concentrated 12 M hydrochloric acid is needed to prepare |

|100.0 mL of a 2.0 M solution? |

|V1 = C2V2 |

|C1 |

|C1=12 M -concentrated |

|C2=2.0 M-diluted |

|V2=100.0 mL-total volume |

| |

|V1 = 2.0 M X 100.0 mL = 17 mL or 0.017 L |

|12 M |

|3) To what volume should 25 mL of 15 M nitric acid be diluted to prepare |

|a 3.0 M solution? |

|V2 = C1V1 |

|C2 |

|C1=15 M-concentrated |

|C2=3.0 M-diluted |

|V1=25 mL |

|V2 = 15 M X 25 mL = 125 mL = 130 mL = 0.13 L |

|3.0 M |

| |

|*T*4) To how much water should 50.0 mL of 12 M hydrochloric acid be |

|added to produce a 4.0 M solution? |

|C1= 12 M-concentrated acid |

|V1= 50.0 mL-amount of concentrated acid |

|C2= 4.0 M-diluted acid |

| |

|V2 = C1V1 |

|C2 |

|V2 = 12 M X 50.0 mL = 150 mL ( Total Volume |

|4.0 M |

|V2 - V1 |

|V(water) = 150 mL - 50.0 mL = 100 mL = 1.0 X 102 mL |

| |

|*T*5) To how much water should 100.0 mL of 18 M sulphuric acid be |

|added to prepare a 1.5 M solution? |

|C1= 18 M-concentrated acid |

|V1= 100.0 mL-amount of concentrated acid |

|C2= 1.5 M-diluted acid |

|V2 = C1V1 |

|C2 |

|V2 = 18 M X 100.0 mL = 1200 mL ( Total Volume |

|1.5 M |

|V2 - V1 |

|V(water) = 1200 mL - 100.0 mL = 1100 mL = 1.1 X 103 mL = 1.1 L |

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