Experiment 11-Freezing Point Depression NAME SECTION 1·lf the sample of glacial

Question

Experiment 11-Freezing Point Depression NAME SECTION 1·lf the sample of glacial acetic acid was old and had absorbed water vapor from the air so that it contained 6.0% water by mass, calculate the freezing point temperature of this solution. [Here, water is the solute. Choose an appropriate mass basis, such as 100g, from which you need to calculate the mass of solute and the mass of solvent.] The normal freezing temperature of glacial acetic acid is 16.0 °C and K 39 Ckg/mol. 2. In the store, you can buy a 50% by mass solution of antifreeze (50% water, 50% ethylene glycol, CrHO Assume that water is the solvent. Calculate the freezing point temperature of the solution. 3. Which of the following aqueous solutions has (have) the lowest freezing point temperature(s)1? Explain 1 m sugar, 1 m sodium nitrate, 1 m aluminum bromide, 1 m potassium sulfate?

Explanation / Answer

1. Let total mass of acetic acid be 100g

6% by mass of water was absorbed hence mass of water = 6g

Actual Mass of glacial acetic acid = 100-6 = 94g

Molality of water in acetic acid = 6×1000/(18×94) = 3.546 molal

?Tf = m × Kf = 3.546 × 3.9 = 12.999 = 13.829°C

Hence depression in freezing point is 13.829°C

Then actual freezing point of acetic acid = 16 - 13.829 = 2.17°C

2. assuming 100g total mass, 50g water + 50g ethylene glycol.

Molality of ethylene glycol in water = (50×1000)/(62×50) = 16.129

?Tf = Kf × m = 1.86(molal freezing constant of water) × 16.129

= 29.99 = 30°C

Freezing point of antifreeze solution = 0-30°C = -30°C

3. Depression in freezing point is a colligative property. Hence it depends only on the no of particles of solute and is independent of the nature or identity of the solute. Hence of the given solutions, AlBr3 will give the higher number of ions (4) hence will have the highest magnitude of freezing point depression, or in other words, lowest freezing point temperature.

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