Question

Question Details:

A student observes that a 126 mL dry flask with a foil capweighs 50.162g. The student then adds 5 mL of an unknown volatileliquid and heats the flask ina boiling water bath at 99.2C(degrees) until all liquid disappears. The flask, foil, andunknown after heating that is dry and cool weigh 50.610 g.Thebarometric pressure in the laboratory is 743.3 mm Hg; thelaboratory temperature averages 22.3 C(degrees); the vapor pressureof the unknown liquid is 160mmHg at 20 C(degrees), and the densityof dry air at STP is 1.2929 g/L.

a) What are the experimental conditions of T and P thatwill be used to calculated the mass of air in the flask?

b) Calculate the density of air at the experimentalconditiosn identified in part a.

c) Calculate the mass of air in the flask before theliquid was added.

d) Calculate the partial pressure of air in the flaskafter the liquid was heated and then cooled.

e) Calculate the mass of air in the flask after the liquidwas heated and then cooled.

f) Calculate the mass of air lost.

g) Calculate the mass of liquid, vapor and air in theflask before you correct for air lost.

h) Calculated the mass of liquid, vapor and air in theflask after you correct for air lost.

i) Calculated the molecular weight of the unknownliquid.

Answer 1

a) T and P at experimental conditions are the pressures andtemperatures you measured in the lab, 743.3 mm Hgand 22.3 C(degrees)respectively.

b) Assuming ideal gas behavior, you can use the ideal gas law tocalculate the density.


743.3mmg0.0404 17-L-mmHg " 0.0404 mol RT X 295.3K raol-A


Now you can convert this to g/L.


niol ,-0.0404-X28.97-01 = 1.17


c) Since you know 126mL of air is in the flask at the beginning,you can calculate its mass using the density.


n,-pl/= 1.17. 0.126L = 0.147g


d) The partial pressure of air is


P unknown 743.3mmHg160mmig583.3mmHg


e) Here you calculate the number of moles of air after adding theliquid


583.3mmHg X 0.1311 62.3637 :mwta × 295.3K air 0.004 15mol RI mol-A


Then you can multiply this by the molecular weight of air to getthe mass

S-0.120g nMW- 0.00415mol X 28.97-s-


f) Just substract the final mass from the initial mass