1. Freezing point of pure acedic acid: 14.5 degrees C volume of water/acedic aci

Question

1. Freezing point of pure acedic acid: 14.5 degrees C volume of water/acedic acid solution: 20 ml acedic acid and 1 ml water freezing point of water/acedic acid solution: 7.9 degrees celsius a.) what is the molality of water in acedic acid solution? b.) what is delta Tf? c.) what is Kf?
2. freezing point of pure acedic acid: 14.5 degrees celsius volume of pure acedic acid in solution: 20 ml mass of unknown solute used in solution: .72 g freezing point of solution of sample: 13.5 degrees celsius a.) what is delta Tf? b.) what is the calculated molality of solution of sample? c.) what is the number of moles of sample in solution? d.) what is the molar mass of sample? 1. Freezing point of pure acedic acid: 14.5 degrees C volume of water/acedic acid solution: 20 ml acedic acid and 1 ml water freezing point of water/acedic acid solution: 7.9 degrees celsius a.) what is the molality of water in acedic acid solution? b.) what is delta Tf? c.) what is Kf?
2. freezing point of pure acedic acid: 14.5 degrees celsius volume of pure acedic acid in solution: 20 ml mass of unknown solute used in solution: .72 g freezing point of solution of sample: 13.5 degrees celsius a.) what is delta Tf? b.) what is the calculated molality of solution of sample? c.) what is the number of moles of sample in solution? d.) what is the molar mass of sample? volume of water/acedic acid solution: 20 ml acedic acid and 1 ml water freezing point of water/acedic acid solution: 7.9 degrees celsius a.) what is the molality of water in acedic acid solution? b.) what is delta Tf? c.) what is Kf?
2. freezing point of pure acedic acid: 14.5 degrees celsius volume of pure acedic acid in solution: 20 ml mass of unknown solute used in solution: .72 g freezing point of solution of sample: 13.5 degrees celsius a.) what is delta Tf? b.) what is the calculated molality of solution of sample? c.) what is the number of moles of sample in solution? d.) what is the molar mass of sample?

Explanation / Answer

Molality = moles of water/ kg of solvent

Assuming density of water = 1 g/cc, mass of water= 1*1= 1 gm, moles of water =1/18= 0.056

Density of acetic acid = 1.05 g/cc and mass of acetic acid =20*1.05 gm=21 gm =21/1000 =21*10-3 kg

Molality of water = 0.056/21*10-3 =2.64 m

From freezing point depression = kf*m*i

I=1 for acetic acid

Kf= (14.5-7.9)/ 2.64 =2.5 deg.c/m

2.

Delta T = 14.5-13.5 =1 deg.c

Mass of acetic acid =20*1.05 =21 gm

Molality= moles of solute/ kg of solvent

kf= 3.58

delta T= kf*m, m =molality

m= 1/3.58 =0.28 moles/ kg solvent

1000 gm of solvent contains 0.28 moles of sample

21 gm of solvent contains 21*0.28/1000 =0.00588 moles

Moles= mass/ Molecular weight

0.00588= 72/ Molecular weight

Molecular weight= 72/0.00588=12245

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