all three please Student Name: EXP #3: PRE-LAB EXERCISE 1. Using Instructor Name

Question

all three please

Student Name: EXP #3: PRE-LAB EXERCISE 1. Using Instructor Name: the information from pages 1-3, calculate the mass of ethylene glycol C H(OH) dissolved in 50.509 g water which would produce a solution a boiling point which is 3.0 degrees Celsius higher than pure water. 2. Which would depress the freezing point more, 10.0 grams of KCl, or 10.0 grams of CaCl2? Explain 3. If a non-volatile solute is dissolved in water, which temperature change would be greater, the boiling point elevation, or the frezing point depcesion? Explain

Explanation / Answer

1. Elevation in boiling point

dTb = Kb.m

with,

dTb = 3 oC

Kb = 0.512 oC/m for water

mass of solvent water = 50.509 g = 0.050509 kg

molar mass ethylene glycol = 62.07 g/mol

m = molality of solution = mass of ethylene glycol (g)/62.07 g/mol x 0.050509 kg

So,

3 = 0.512 x mass ethylene glycol/62.07 x 0.050509

mass of ethylene glycol = 18.37 g required

2. 10 g CaCl2 with ions in solution = 3 (1 Ca2+ and 2 Cl-) would dress freezing point more than KCl with two ions in solution (K+ and Cl-).

3. The non-volatile solute would cause larger depression in freezing point than the elevation in boiling point upon its addition to water. This is because more heat is needed to increase boiling point and to break the intramolecular bonds in water than forming bonds when water is frozen.

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