The first part looks ok to me.
The second part (i for KCl): You haven't shown any difference between the two methods.
delta T = i*Kf*m
i = delta T/K*m
i = 5.00/1.86*0.1 = 26.9 so your math for your first calculation is ok (but of course the correrct answer is nowhere near 26.9). The second calculation is EXACTLY the same set up so I don't know how you arrived at 0.2688.
Hey guys I did a lab and I have to answer the questions relating to the lab. I did question can someone please check if these are accurate
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:
FOR PART 2 I AM NOT SURE WHAT IS THE RIGHT METHOD OF DOING IT, I DID IT TWO WAYS CAN SOMEONE PLEASE CHECK, WHICH IS THE RIGHT ONE.
Kf = 1.86°C kg/mole
Experimentally determined freezing point of KCl: -5.00
m KCl (given): 0.1 m (molality)
1. . Solution: i
= -5.00°C /(1.86°C kg/mole)(0.1)
i = 26.88
or
2. -5.00°C /-1.86°C kg/mole * 0.1m KCl
i = 0.2688
So is i 0.2688 or would it be 26.88
4 answers
1. deltaTf = 0 - (-1.0) = 1.0 deg
Kf = 1.86
m = 1.0deg/1.86deg/molal unit
Final answer should be OK
2. i = (molality calculated from your experiemental deltaTf) / (molality calculated from moles KCl and kilograms of water)
Your answer (value of i) should be between 1 and 2, probably closer to 2
Kf = 1.86
m = 1.0deg/1.86deg/molal unit
Final answer should be OK
2. i = (molality calculated from your experiemental deltaTf) / (molality calculated from moles KCl and kilograms of water)
Your answer (value of i) should be between 1 and 2, probably closer to 2
Additional comment:
1 mole of KCl produces 2 moles of particles (ions). Because of this, a 1.00m solution of KCl would behave like a 2.00m solution as far as freezinf point, boiling point, and other colligative properties are concerned. That assumes 100% dissociation of KCl ionto ions in solution. The FPD would be (2.0)(1.86)=3.72) and the freezing point would be 0-3.72 = -3.72 deg.C. The actual freezing point lowering is less than predicted (assuming 100% dissociation) because some "ion pairs" behave like a single particle. Your 0.1m solution of KCl should produce the FPD of a 0.2m solution (-0.2*1.86=-0.376). Your definition, i = deltaT/Kf.m, is ok but it is producing an answer that is not reasonable. Probably poor experimental data.
(molality in terms of moles of ions) divided by (molality in terms of formula units)should give something less than (2 ions)/(formula unit).
1 mole of KCl produces 2 moles of particles (ions). Because of this, a 1.00m solution of KCl would behave like a 2.00m solution as far as freezinf point, boiling point, and other colligative properties are concerned. That assumes 100% dissociation of KCl ionto ions in solution. The FPD would be (2.0)(1.86)=3.72) and the freezing point would be 0-3.72 = -3.72 deg.C. The actual freezing point lowering is less than predicted (assuming 100% dissociation) because some "ion pairs" behave like a single particle. Your 0.1m solution of KCl should produce the FPD of a 0.2m solution (-0.2*1.86=-0.376). Your definition, i = deltaT/Kf.m, is ok but it is producing an answer that is not reasonable. Probably poor experimental data.
(molality in terms of moles of ions) divided by (molality in terms of formula units)should give something less than (2 ions)/(formula unit).
thank you
3 answers