Ex 10

‫بسم ال الرحمن الرحيم‬ Name: ayedzahran Group: Day:

Experiment 10 "Molecular weight of a volatile liquid" _________________________________________________________________ I. Observations and results: Mass of dry apparatus Mass of apparatus and condensed vapor Mass of condensed vapor Total volume of flask (ml) Total volume of flask (L) Barometric pressure (torr) Barometric pressure (atm) Temperature of vapor (Cº) Temperature of vapor (K) Molecular weight of liquid (g\mol) II. Calculation: (1) Mass of condensed vapor = 92.3-91.9=0.4 g (2) Total volume of flask (L)=338*10-3 =0.338 L

(3) Barometric pressure =750/760=0.987 atm (4) Temperature of vapor (K)= 98+273=371 K (5) P*V=n*R*T ,n=m/M M = (m*R*T)/(P*V) = (0.4*0.082*371)/(0.987*0.338) = 36.48 g\mol

First trial 91.9 g 92.3 g 0.4 g(1) 338 ml 0.338 L(2) 750 torr 0.987 atm(3) 98 Cº 371 K(4) 36.48 g\mol(5)

III. Answer the following questions: 1. How would your experiment results be effected if : a. Some water was left on the apparatus The water will affect weight of the flask and the results will be in correct. Molecular weight will increase. b. The liquid not completely evaporated. This gives in correct calculations since when we scale the apparatus to measure the weight of vapor in the flask. This evaporated liquid will measured with vapor and this increase the molecular weight obtained from the calculations c. You forgot to make a pinhole in the foil. The pressure inside the flask will increase and being larger than atmospheric pressure ,and any liquid that required to fill the flask as vapor can’t escape as liquid is heated so this will incorrect results (decrease the molecular weight). 2. If 0.80 g of the vapor of unknown liquid occupied 280 ml at 100 C° and

750 torr, calculate the molecular weight of the liquid. P= 750 torr = 0.987 atm, T=100 C = 373 K, V=280 ml=0.28 L P*V= n*R*T, n=m/M M=(m*R*T)/(P*V) =(0.8*0.082*373)/(0.987*0.28) = 88.54 g\mol Not: For experiment number , For calculation number , For the solution of answers .