1. Which solution has the lower freezing point, 1.0 M KBr or 1.0 M CaBr? 2. What

Question

1. Which solution has the lower freezing point, 1.0 M KBr or 1.0 M CaBr? 2. What would be the boliling points of the solutions in Exercise 1? 3. Which solution has the higher boiling point, 0.75 M glucose or 0.6 M K CI? 4. Antifreeze in cars can help prevent the engine from overheating in warm climates. If e glycol mponent, how many grams of eth n co would need to be added to 2 L of water to raise the boiling point to 110°? 5. Rock salt (NaCl) is added to ice to provide a low enough temperature to make homemad temperature of the ice/water mixture to -20°C? e ice cream. What percent Naci solution needs to be achieved to lower the 6. Read the ässigned pages i In your textbook, and work the assigned problems.

Explanation / Answer

use the formula

Delta T = i * m * kb

where i is the vant hoff factor

m is molality

kf is the freezing point depression constant

for water it is 1.86

We need to change from molarity to molality for pure water (density = 1 g/ml)

Kbr will dissociate in water into K+ and Br-

2 ions so vant hoff factor is 2

Delta T = 1.86 * 2 * 1 = 3.72 C

Delta T = Tf pure solvent - Tf solution

3.72 = 0 - Tf solution

Tf solution = 0 - 3.72 = -3.72 C

For CaBr2

it will dissociate into 3 ions, 1 Ca+ and 2 Br-1

vant hoff factor is 3

Delta T = 3 * 1.86 * 1 = 5.58 C

100 - T = 5.58C

T = 0 - 5.58 = -5.58 C

Question 2

is the same but instead of 1.86 use 0.52 so

KBr

Delta T = 2 * 1 * 0.52 = 1.04 C

Delta T = Tb solution - Tb pure solvent

Tb solution - 100 = 1.04

Tb solution = 100 + 1.04 = 101.04 C

For CaBr2

vant hoff factor is 3

deltaT = 3 * 1 * 0.52 = 1.56

Tb solution = 100 + 1.56 = 101.56 C

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