lab simulation flashcards - colligative properties lab quiz

What is molality?

Molality (m) = moles of solute / kilograms of solvent.

Why is molality used in colligative properties instead of molarity?

Molality is independent of temperature because it’s based on mass, not volume.

How do you calculate molality from mass and molar mass?

Find moles of solute (g / molar mass), then divide by kg of solvent.

Example: Molality of 125g glucose in 500g water

125g / 180.16g/mol = 0.694 mol; 0.694 mol / 0.5 kg = 1.39 m

How do you calculate molality with volume and density of solvent?

Convert volume to mass using density, then divide moles of solute by solvent mass in kg.

Example: Molality of 78g butanone in 800mL acetic acid

78g / 72.11g/mol = 1.082 mol; 800mL x 1.049g/mL = 839.2g = 0.8392 kg; m = 1.29 m

How do you find molar mass from freezing point depression?

Use ΔTf = Kf × m, find molality, then calculate moles and divide mass / moles.

Example: Molar mass with ΔTf = 2.34°C and Kf = 1.86

ΔTf/Kf = 1.26 m; 93.24g / 1.26 mol = 74.0 g/mol

What is the formula for freezing point depression?

ΔTf = i × Kf × m

What is Kf for acetic acid?

3.90 °C·kg/mol

What is the freezing point of pure acetic acid (HAc)?

15.80 °C

How do you calculate mass of solvent in kg from volume and density?

Mass = volume × density; Convert to kg

What is the ΔTf for solution #1 in acetic acid lab?

15.80°C - (-3.5°C) = 19.3°C

What was the molality for solution #1?

ΔTf / Kf = 19.3 / 3.90 = 4.95 m

How many moles of solute in solution #1?

m × kg solvent = 4.95 × 0.03147 = 0.1558 mol

What is the molar mass of unknown in solution #1?

1.186g / 0.1558 mol = 7.62 g/mol

What was ΔTf for solution #2?

15.80°C - (-6.5°C) = 22.3°C

What was molality for solution #2?

22.3 / 3.90 = 5.72 m

Moles of solute in solution #2?

5.72 × 0.03147 = 0.1801 mol

Molar mass of unknown in solution #2?

2.373g / 0.1801 mol = 13.18 g/mol

Average molar mass of unknown?

(7.62 + 13.18) / 2 = 10.40 g/mol

What was the calculated Kf from 0.5 m sucrose solution?

13.1°C / 0.5 m = 26.2 °C/m

What was the calculated Kf from 1.0 m sucrose solution?

11.40°C / 1.0 m = 11.40 °C/m

Average calculated Kf for water?

(26.2 + 11.4) / 2 = 18.8 °C/m

How accurate were your Kf results for water?

Not accurate. % error = |18.8 - 1.86| / 1.86 × 100 = 911%

What was i (van’t Hoff factor) calculated for sucrose?

i = 1 (sucrose is non-electrolyte)

What’s the significance of measuring freezing point accurately?

Small temp errors cause large changes in molar mass calculations.

New molar mass with Tf = 11.0°C in follow-up?

ΔTf = 5.1; m = 1.31; moles = 1.31 × 0.0315 = 0.0413 mol; molar mass = 3 / 0.0413 = 72.64 g/mol

What does freezing point depression show about rock salt and ice?

Ions disrupt ice structure, cause it to melt, absorbing heat and lowering temp below 0°C.

Major sources of error in the experiment?

Imprecise temp readings, poor mixing, contamination, supercooling.

How do you improve accuracy in colligative property labs?

Use calibrated thermometer, ensure uniform mixing, avoid supercooling, repeat trials.