Scientific Understanding

**What are freezing point depression and boiling point elevation?**

Freezing point depression and boiling point elevation are talking about how you take a solution and influence it to change its freezing point (freezing temperature) or boiling point (boiling temperature). The freezing point and the boiling point change when you add more of either the solute or the solvent to the solution therefore, changing it’s molality. The vast majority of the time this happens you will cause the freezing point to decrease (depress) and the boiling point to increase (elevate). Books and teachers usually talk about this in terms of adding or changing only the solute because it is more simple to explain if only one of the factors is changing.

**Where is ****freezing point depression and boiling point elevation ****used in every day life?**

Most people tend to use these concepts to improve their lives at least once a week. If you are cooking noodles and you want them to cook faster, then you throw some salt in the water that you are boiling and this raises (elevates) the boiling point of the water and therefore the noodles cook faster. If you live in a cold climate, then when it snows trucks come around to spread salt on the roads. This salt mixes with the ice or water and causes the freezing point to decrease (depress) and therefore the ice can more easily turn back into the liquid form of water and run off the street to decrease slips and falls and car accidents.

**What is the formula for freezing point depression and boiling point elevation?**

The formula for freezing point depression and boiling point elevation is below.

Δ T = |
m * k * i |

1 |

Delta T ( **Δ** T ) represents the change in temperature with units of degrees Celsius. Little m represents molality with units of mols per kg (mol / kg). The k represents the constant for the change in freezing point (k_{f}) or the constant for the change in boiling point (k_{b}). Each chemical you deal with has a different k_{f} or k_{b} and you only use the k_{f} or k_{b} of the chemical that is the solvent. For this learning section I am only going to use the k_{f} and k_{b} of water (H_{2}O) since the most common solvent that most teachers use especially in high school. Finally we come to the I which represents the ionization constant that we discussed in the learning section before this one. Let us check out the examples below.

For water:

K_{f} = -1.86 C kg / mol

K_{b} = 0.512 C kg / mol

**VIDEO Freezing Point Depression and Boiling Point Elevation ****Demonstrated Example ****1**: If you put 2 moles of NaBr into 200 kg of water, then what will be your change in freezing point? K_{f} = -1.86 C kg / mol

**Step 1: ****Gather information**

mol of NaBr = 2 moles

mass of water = 200kg

K_{f} = -1.86 C kg / mol

i is related to NaBr

i = ?

m = ?

**Δ** T = ?

**Step ****2****: ****which equation connects my information?**

**Δ** T = m * k * i

**Step ****3****: ****solve for i**

Since i is related to the chemical NaBr then when we put NaBr in water it breaks up since it is ionic. It breaks up into two ions Na^{+1} and Br^{-1}.

Answer: i = 2

**Step ****4****: ****solve for m**

The m or molality = mol solute / kg of solvent.

m = | mol solute |

kg solvent |

**Step ****5****:**

m = | 2 mol NaBr |

200 kg water |

**Step ****6****: ****calculate**

m = | 2 mol |

200 kg |

Answer: m = 0.01 mol / kg

**Step ****7****: ****solve for ****Δ**** ****T**

Δ T = |
m * k * i |

1 |

**Step ****8****: ****put in numbers and units**

Δ T = |
0.01 mol * -1.86 C kg * 2 |

kg mol |

**Step ****9****: ****cross out units**

Δ T = |
0.01 mol * -1.86 C kg * 2 |

kg mol |

**Step ****10****: ****Simplify**

Δ T = |
0.01* -1.86 C * 2 |

1 |

**Step ****1****1****: ****calculate**

0.01* -1.86 * 2 = – 0.0372

COMPLETE ANSWER: **Δ** T = – 0.0372

**VIDEO Freezing Point Depression and Boiling Point Elevation ****Demonstrated Example ****2**: If the change in temperature for the boiling point is 2.1 C with 500 L of water then how many moles of NF_{3} did you put into the water? K_{b} = 0.512 C kg / mol

**Step 1: ****Gather information**

**Δ** T = 2.1 C

mass of water = 500 L = 500 kg

K_{b} = 0.512 C kg / mol

entire formula is **Δ** T = m * k * i

i is related to NF_{3}

i = ?

m = ?

mols of NF_{3} = ?

**Step ****2****: ****which equation connects my information?**

**Δ** T = m * k * i

**Step ****3****: ****solve for i**

Since i is related to the chemical NF_{3} then when we put NF_{3} in water it does not break up since it is covalent. Therefore, it remains as one molecule.

Answer: i = 1

**Step ****4****: ****solve for m**

Δ T = |
m * k * i |

1 | 11 |

**Step ****5****: ****put in numbers and units**

2.1 C = | m * 0.512 C kg / mol * 1 |

1 |

**Step ****6****: ****divide both sides by the K**_{b}** and the i**

2.1 C = | m * 0.512 C kg / mol * 1 |

0.512 C kg / mol * 1 | 0.512 C kg / mol * 1 |

**Step ****7****: ****cross out**** the ****matching stuff on the right side.**

2.1 C = | m * 0.512 C kg / mol * 1 |

0.512 C kg / mol * 1 | 0.512 C kg / mol * 1 |

**Step ****8****: ****simplify**

The m or molality = mol solute / kg of solvent.

2.1 C = | m |

0.512 C kg / mol * 1 |

**Step ****9****: ****calculate**

2.1 C / ( 0.512 C kg / mol * 1 ) = 4.10

m = 4.10 mol / kg

**Step ****10****: ****solve for the moles of NF**_{3}** **

The m or molality = mol solute / kg of solvent.

m = | mol solute |

kg solvent |

**Step ****1****1****: ****fill in the numbers and units**** **

4.10 mol / kg = | mol solute |

500 kg |

**Step ****1****2****: ****multiply both sides by 500 kg**

500 kg * 4.10 mol / kg = | mol solute * 500 kg |

500 kg |

**Step ****1****3****: ****cross out the 500kg on the right side**

500 kg * 4.10 mol / kg = | mol solute * 500 kg |

500 kg |

**Step ****1****4****: ****simplify**

500 kg * 4.10 mol / kg = | mol solute |

1 |

**Step ****1****5****: ****calculate**

500 * 4.10 = 2050

COMPLETE ANSWER: moles of NF_{3} = 2050 moles

**PRACTICE PROBLEMS**: Solve the freezing point depression and boiling point elevation problems below. Try to use a regular periodic table. However if you need you can use your metal / non-metal periodic table and your ion periodic table. Remember the constants for water…….K_{f} = -1.86 C kg / mol……and..….K_{b} = 0.512 C kg / mol

If you put 3.0 moles of LiBr into 150 kg of water what will be your change in boiling point?

Answer: **Δ** T = 0.020 C

If you put 5.0 moles of Br_{2} into 120kg of water what will be the change in freezing point?

Answer: **Δ** T = – 0.078 C

QUESTION SOLVING for molality

How many moles of CaF_{2} will it take to change the boiling temperature of a solution with 305 L of water by 1.3 degrees C?

Answer: 260 moles

If an unknown covalent chemical changes the freezing temperature of a solution of 40 kg of water by -0.60 C then how many moles of that covalent chemical were added?

Answer: 12.9 moles