Posted by Sev on Friday, February 6, 2009 at 3:36pm.
Hey guys I did a lab and I have to answer the questions relating to the lab. I did question 1 & 2, can someone check if they are correct, and question 3 I don't know how to do it.
1. Calculation to Determine the molecular weight of unknown substance
Mass of unknown used: 2.0 g
Mass of water used: 50.0 g (0.05kg)
Kf = 1.86°C kg/mole
Experimentally determined freezing point of unknown: -1°C
Solution = delta Tf/-Kf
= -1°C/-1.86°C kg/mole
= 0.538 m (molality)
Molar mass = 2g / 0.05 kg * 0.538 m
= 74.4 g/mol
2. Calculations to determine i for KCl solution:
Kf = 1.86°C kg/mole
Experimentally determined freezing point of KCl: -1°C
m KCl (given): 0.1 m (molality)
Solution: i = measured delta Tf/ expected delta Tf
= -1°C /-1.86°C kg/mole * 0.1m KCl
i = 5.376
3. Explain why i has this value = ??
chemistry, freezing point depression - bobpursley, Friday, February 6, 2009 at 3:51pm
The first is correct IF the substance was non ionizing.
The thing that strikes me is precision. You measured the freezing point to ONE significant digit, and then did math with it. OUCH.
What if your measured deltaT in the second had been 1.4999, which is in tolerance for your 1 +-.5C measurement.
If it had been 1.49999, then i would have been 5.8, but if delta T had been .5, then i would have been 2.6 This is quite a range, .8 to 2.6. Frankly, I would have expected i to be 2.0 for KCl, which breaks into two ions.
So I attribute your results to lack of precision in measurements.
chemistry, freezing point depression - Anonymous, Saturday, February 7, 2009 at 2:24pm
Do impurities take into account? for #3?
chemistry, freezing point depression - ur class fella, Sunday, February 8, 2009 at 10:59pm
yeah it should be around or less den 2.
i mean the i should be = 2 because it breaks into to ions for KCL and i am getting 1.6. pretty close eh.
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