3, A student observes that a 126 mL dry flask with a foil cap weighs 50.162 g. T

Question

3, A student observes that a 126 mL dry flask with a foil cap weighs 50.162 g. The student then adds 5 mL of an unknown volatile liquid and heats the flask in a boiling water bath at 99.2 C until all liquid disappears. The flask, foil, and unknown after heating that is dry and cool weigh 50.610 g. The barometric pressure in the laboratory is 743.3 mm Hg; the laboratory temperature averages 22.3°C; the vapor pressure of the unknown liquid is 160 mmHg at 20°C, and the density of dry air at STP is 1.2929 g/L. a) What are the experimental conditions of T and P that will be used to calculate the mass of air in the flask? PV 43m.iao L) AT(a2 +373 ) b) Calculate the density of air at the experimental conditions identified in part a 1.807a glmL c) Calculate the mass of air in the flask before the liquid was added. d) Calculate the partial pressure of air in the flask after the liquid was heated and then cooled. e) Calculate the mass of air in the flask after the liquid was heated and then cooled. f) Calculate the mass of air lost. g) Calculate the mass of liquid, vapor and air in the flask before you correct for air lost. h) Calculate the mass of liquid, vapor and air in the flask after you correct for air lost. i) Calculate the molecular weight of the unknown liquid.

Explanation / Answer

Answering C) to i), expect answers g) and h)

c) Mass of air in the flask before addition of Liquid :

first we will calculate the no of moles of Gas in the flask from ideal gas eq. PV = n RT

n = PV/RT = (743.4/760 atm) X (0.126 L) / (0.082 L atm K-1 mol-1* (22.3+273 K) = 0.005 moles

Since mol mass of the air = 28.95 gm/mole

Mass of air = 0.005 moles x 28.95 gm/mole = 0.144 gms.

d) Since the flask is open the Partial pressure of the air will be equal to the Atmospheric pressure (barometer reading) i.e 743.3 mm Hg.

e) Mass of air in the flask when liquid is heated and cooled:

In order to determine how many moles of air remaining once the volatile liquid and vapor reach dynamic equilibrium, the combined gas law must be applied to determine what the volume of the vapor would have been at room temperature.

P1V1/T1 = P2V2/T2 = (743.4/760 atm) X (0.126 L)/ (99.2+273 K) = (743.4+160/760 ATM) x V2 / (22.3+273 K)

V2 = 0.082 L (Volume of Vapor)

This means that .126 L .082 L = .044 L, of air remained in the flask with the vapor and liquid. In order to determine the mass of this air, PV=nRT must be used,

(743.4/760 atm) X (0.044 L) / (0.082 L atm K-1 mol-1* (22.3+273 K) = 0.00176 moles

Since mol mass of the air = 28.95 gm/mole

Mass of air = 0.00176 moles x 28.95 gm/mole = 0.051 gms.

f) Mass of Air lost = Mass of air initial - mass of air when liquid is heated and cooled = 0.144 gms - 0.051 gms.= 0.093 gm

i) Mol wt of the unknown liquid :

There were .0.051 grams of air in the flask with the vapor. This can be subtracted from the observed amount to find out the actual amount of liquid condensed.

50.610 gm - 50.162 - 0.051= 0.397 gm

There were 0.397 gm of unknown liquid in the flask, taking the air in the flask into account.

The number of moles of vapor can be caluted by

PV = n RT

n = PV/RT ( T = 373 K, V = 0.126 L, )

= (743.4+160/760 ATM) x 0.126 L / (0.082 L atm K-1 mol-1)X (373 K)

= 1.18 x 0.126/ 30.58 = .004 mols

Mol wt of the unknown liquid = 0.397 gm / 0.004 mols = 99.25

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