1-Butanol, 1-butene, and di-n-butyl ether are all removed from the product by washing with concentrated sulfuric acid. Explain w
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Answer :
(a) The pressure of the vapor in the flask in atm is, 0.989 atm
(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K
The volume of the flask in liters is, 0.2481 L
(c) The mass of vapor present in the flask was, 0.257 g
(d) The number of moles of vapor present are 0.00802 mole.
(e) The mass of one mole of vapor is 32.0 g/mole
Explanation : Given,
Mass of empty flask and stopper = 55.844 g
Volume of liquid = 5 mL
Temperature =
Mass of flask and condensed vapor = 56.101 g
Volume of flask = 248.1 mL
Barometric pressure in the laboratory = 752 mmHg
(a) First we have to determine the pressure of the vapor in the flask in atm.
Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg
Conversion used :
or,
As,
So,
Thus, the pressure of the vapor in the flask in atm is, 0.989 atm
(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.
Conversion used :
As,
So,
Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K
Now we have to determine the volume of the flask in liters.
Conversion used :
1 L = 1000 mL
or,
1 mL = 0.001 L
As, 1 mL = 0.001 L
So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L
Thus, the volume of the flask in liters is, 0.2481 L
(c) Now we have to determine the mass of vapor that was present in the flask.
Mass of flask and condensed vapor = 56.101 g
Mass of empty flask and stopper = 55.844 g
Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper
Mass of vapor in flask = 56.101 g - 55.844 g
Mass of vapor in flask = 0.257 g
Thus, the mass of vapor present in the flask was, 0.257 g
(d) Now we have to determine the number of moles of vapor present.
Using ideal gas equation:
PV = nRT
where,
P = Pressure of vapor = 0.989 atm
V = Volume of vapor = 0.2481 L
n = number of moles of vapor = ?
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of vapor = 372.7 K
Putting values in above equation, we get:
Thus, the number of moles of vapor present are 0.00802 mole.
(e) Now we have to determine the mass of one mole of vapor.
Thus, the mass of one mole of vapor is 32.0 g/mole