dent was gien a 10 mL sample of a clear, colorless liquid. She was assigned the
ID: 554636 • Letter: D
Question
dent was gien a 10 mL sample of a clear, colorless liquid. She was assigned the task of identiftying the unlnown liquid and was told that the sample could be methanol (CHhOH), acetone CaiO), or ethanol (CaHOH), She decided to attempt to determine the molar mass of the liquid by the vapor density method, which involves completely vaporizing a small sample of the liquid, cooling it and detcrmining the mass of the condensed vapor. She also collects the volume of the container, temperatute and pressure whan the liquid is vaporized. The following data were collected Trial #1 Trial #2 . Mass of empty flask + foil cap Mass of flask, cap, condensed vapor Mass of condensed vapor 125.194 g125.194g 125.470 g 125.592 g 1996C 99.8 Temperature of boiling wat rbathCC) Temperature of boiling water bathK) Atmospheric pressure (um Hg) Atmosphcric pressure (alm) Volume of flast (L Volume of flask (L) 752 mm Hg 756 mm Hg 288 mL 288 mL. 0.288 L 0.288 L (a) Fill in the missing data in the table. (b) Determine the molar masses for each trial, showing all calculations. (c) The student reported the unknown liquid was definitely methanol. Do you agree with her conclusion? Why or why not? 2. A rigid 8.20 L flask contains a mixture of 2.50 moles of H2, 0.500 mole of Os, and sufficient Ar so that the partial pressure of Ar in the flask is 2.00 atm. The temperature is 127°C (a) Calculate the total pressure in the flask. (b) Calculate the mole fraction of H2 in the flask (c) Calculate the density (in g L-1) of the mixture in the flask.Explanation / Answer
1)
Trial #1
Trial #2
A) Mass of empty flask + foil cap
125.194 g
125.194 g
B) Mass of flask, coil and condensed vapor
125.470 g
125.592 g
C) Mass of condensed vapor = B – A
(125.470 – 125.194) g = 0.276 g
(125.592 – 125.194) g = 0.398 g
D) Temperature of boiling water bath (°C)
99.8
99.6
E) Temperature of boiling water bath (K) = (D + 273)
(99.8 + 273) = 372.8
372.6
F) Atmospheric pressure (mmHg)
752
756
G) Atmospheric pressure (atm) = (F)*(1 atm/760 mmHg)
(752 mmHg)*(1 atm/760 mmHg) = 0.9895 atm
(756 mmHg)*(1 atm/760 mmHg) = 0.9947 atm
H) Volume of flask (mL)
288
288
I) Volume of flask (L) = (H)*(1 L/1000 mL)
0.288
0.288
Find out the number of moles of the condensed vapor using the ideal gas law as below.
n = P*V/RT where P = pressure in atmosphere; V = volume in L; T = temperature in Kelvin scale and R = 0.082 L-atm/mol.K is the gas constant. Hence, find the molecular mass of the condensed vapor.
Trial #1
Trial #2
n = P*V/RT
(0.9895 atm)*(0.288 L)/(0.082 L-atm/mol.K).(372.8 K) = 0.009322 mole
(0.9947 atm)*(0.288 L)/(0.082 L-atm/mol.K).(372.6 K) = 0.009376 mole
MW of condensed vapor = (mass of condensed vapor)/n
(0.276 g)/(0.009322 mole) = 29.6074 g/mol 29.61 g/mol
(0.398 g)/(0.009376 mole) = 42.4488 g/mol 42.45 g/mol
Find out the molar masses of the three volatile compounds given.
Methanol, CH3OH = (1*12.01 + 4*1.008 + 1*15.9994) g/mol = 32.0414 g/mol.
Acetone, C3H6O = (3*12.01 + 6*1.008 + 1*15.9994) g/mol = 58.0774 g/mol.
Ethanol, C2H6O = (2*12.01 + 6*1.008 + 1*15.9994) g/mol = 46.0674 g/mol.
The student obtained a result close to the molar mass of methanol in Trial #1 while the MW obtained in Trial #2 is close to the molar mass of ethanol Hence, it cannot be said with certainty whether the unknown liquid is methanol or ethanol. A third trial is necessary to ascertain the exact identity.
Trial #1
Trial #2
A) Mass of empty flask + foil cap
125.194 g
125.194 g
B) Mass of flask, coil and condensed vapor
125.470 g
125.592 g
C) Mass of condensed vapor = B – A
(125.470 – 125.194) g = 0.276 g
(125.592 – 125.194) g = 0.398 g
D) Temperature of boiling water bath (°C)
99.8
99.6
E) Temperature of boiling water bath (K) = (D + 273)
(99.8 + 273) = 372.8
372.6
F) Atmospheric pressure (mmHg)
752
756
G) Atmospheric pressure (atm) = (F)*(1 atm/760 mmHg)
(752 mmHg)*(1 atm/760 mmHg) = 0.9895 atm
(756 mmHg)*(1 atm/760 mmHg) = 0.9947 atm
H) Volume of flask (mL)
288
288
I) Volume of flask (L) = (H)*(1 L/1000 mL)
0.288
0.288
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