A student completed this lab and got the following dat for a liquid unknown.. In

Question

A student completed this lab and got the following dat for a liquid unknown.. Information Given: Lowest observed temperature in ice-water mixture: 0.9 C Mass of liquid unknown added: 11.9 g Trial #1: lowest observed temperature in ice solution mixture: -3.1 C Trial #1: Mass of the solution: 90.4 g Trial #2: Lowest observed temperature in ice-solution mixture: -2.3 C Trial #2: Mass of the solution: 122.9 g A. What is the freezing point depression for trial #1 and trial #2? B. Using the freezing point depression for trial #1 and trial #2, calculate the molality of the solution. C.What is the mass of the water used in trial #1 and #2? D. What is the number of mols of unknown in trial #1 and trial #2.

Explanation / Answer

Freezing point depression =0.9+3.1= 4 deg.c, ( Trial-1) and Freezing point depression = 0.9+2.3 =3.2 deg.c

for water Kf= 1.86deg,c/m

So Freezing point depression= Kf*m

4= 1.86*m

m= 4/1.86 ( trial-1) =2,15 and 3.2/1.86=1.72 ( Trial-2), Average molality= (2.15+1.72)/2=1.935

Mass of liquid added= 11.9 gms, mass of water =90.4-11.9=78.5 gms kg =78.5/1000 =0.0785 kg( Trial-1 ) and 122.9-11.9=111gm=111/1000=0.111 kg

molality= moles of soute/ kg of water. moles of solute= molality* kg water

trial-1 moles of solute= 2.15*0.0785=0.1687, Trial- moles of solute= 1.72*0.111=0.190

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