Molarity Problems Part Deux

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- Molarity Calculations

Calculate the molarities of the following solutions:

**1) 2.3 moles of sodium chloride in 0.45 liters of water.**

(2.3 moles) / (.45 liters) = 5.1 moles/liter

**2) 1.2 moles of calcium carbonate in 1.22 liters of water.**

(1.2 moles) / (1.22 liters) = .984 moles/liter

**3) 0.09 moles of sodium sulfate in 12 mL of water.**

(0.09 moles) / (.012 liters) = 7.5 moles/liter

**4) 0.75 moles of lithium fluoride in 65 mL of water.**

(.75 moles) / (.064 liters) = 11.7 moles/liter

**5) 0.8 moles of magnesium acetate in 5 liters of water.**

(.8 moles) / (5 liters) = .16 moles/liter

**6) 120 grams of calcium nitrite in 240 mL of water.**

**Atomic mass of Ca(NO**_{2})_{2}= 20.08 g/mole + (14.01 g/mole)(2) + (16.00 g/mole)(4) = 112.1 g/mole**(120 g)/(112.1 g/mole) = 1.07 moles****(1.07 moles) / (.240 liters) = 4.46 moles/liter**

7) 98 grams of sodium hydroxide in 2.2 liters of water.

8) 1.2 grams of hydrochloric acid in 25 mL of water.

9) 45 grams of ammonia in 0.75 L of water.

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**Explain how you would make the following solutions. You should tell how many
grams of the substance you need to make the solution, not how many moles.**

**10) 2 L of 6 M HCl**

**1 liter of a 6 M solution means there are 6 moles of HCl per liter, so...****the atomic mass of HCl is (1.008 g/mole) + (35.45 g/mole) = 36.45 g/mole****1 mole of HCl is: (1 mole)(36.45 g/mole) = 36.45 grams****However, we don't want 1 mole, we want 6 moles, so we increase our***recipe*by 6: 6 x 36.45 grams = 218.7 g**We're still not done though, we need twice as much solution (2 liters) so we increase our recipe by 2: (218.7g)(2) = 437.4g**

**Railroad Tracks:**

6.0 moles HCl | 36.45 g HCl | 2 liters |
---|---|---|

1 liter |
1 mole HCl |

**= 437.4 grams HCl ( ack sig figs are messed up here...)**

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**11) 1.5 L of 2 M NaOH**

**Let's keep it simple and make 1 liter first so we need 2 moles / liter****NaOH = (22.99 g/mole) + (16.00 g/mole) + (1.008 g/mole) = 40.00 g/mole****We want 2 moles so (40.00 g/mole)(2 mole) = 80.00 grams****So to make 1 liter of 2M NaOH takes 80.00 grams so to make 1.5 liter we would need 1.5 times as much NaOH ==> 80.00 grams x 1.5 = 120.0 grams**

**Railroad tracks**

2 moles NaOH | 40.00 g NaOH | 1.5 liters |
---|---|---|

1 liter |
1 mole NaOH |

**=120 grams NaOH**

**12) 0.75 L of 0.25 M Na _{2}SO_{4}**

**We want a .25 M solution, so let's make 1 liter first****We need the atomic mass first: (22.99 g/mole)(2) + (32.07 g/mole) + (16.00 g/mole)(4) = 142.1 g/mole****(.25 moles)(142.1 grams/mole) = 35.51 grams****However, that is for 1 liter of solution and we want .75 liters so...****(.75)(35.51 grams) = 26.63 grams**

.25 moles Na_{2}SO_{4} |
142.1 g Na_{2}SO_{4} |
.75 liters |
---|---|---|

1 liter |
1 mole Na_{2}SO_{4} |

**= 26.64 grams** **of** **Na _{2}SO_{4}**

**13) 45 mL of 0.12 M sodium carbonate**

.12 moles Na(CO_{3}) |
83.00 g Na(CO_{3}) |
1 liter_{} |
45 ml |
---|---|---|---|

1 liter |
1 mole Na(CO_{3}) |
1000 ml_{} |

**=.4482 grams of Na(CO _{3})**

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**14) 250 mL of 0.75 M lithium nitrite**

.75 moles Li(NO_{2}) |
36.95 g Li(NO_{2}) |
1 liter | 250 ml |
---|---|---|---|

1 liter |
1 mole Li(NO_{2}) |
1000 ml |

**= 6.93 grams Li(NO _{2})**

15) 56 mL of 1.1 M iron (II) phosphate

16) 6.7 L of 4.5 M ammonium nitrate

17) 4.5 mL of 0.05 M magnesium sulfate

18) 90 mL of 1.2 M BF3