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September 24, 2016
Posted by **Saira** on Wednesday, May 27, 2009 at 9:15pm.

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

- Chemistry please help -
**DrBob222**, Wednesday, May 27, 2009 at 9:42pmThe 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. - Chemistry please help -
**GK**, Wednesday, May 27, 2009 at 9:43pm1. 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 - Chemistry please help -
**GK**, Wednesday, May 27, 2009 at 10:36pmAdditional 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). - Chemistry please help -
**Saira**, Wednesday, May 27, 2009 at 11:12pmthank you