6.103. A sample of oxygen was collected over water at 25°C and 1.00 atm. a. If t

Question

6.103. A sample of oxygen was collected over water at 25°C and
1.00 atm.
a. If the total sample volume was 0.480 L, how many moles of O2 were collected?
b. If the same volume of oxygen is collected over ethanol instead of water, does it contain the same number of
moles of O2?
6.104. Water and ethanol were removed from the O2 samples in Problem 6.103.
a. What is the volume of the dry O2 gas sample at 25°C and 1.00 atm?
b. What is the volume of the dry O2 gas sample at 25°C and 1.00 atm if P (ethanol) = 50 mmHg at 25°C ?

Explanation / Answer

Solution :-

6.103. A sample of oxygen was collected over water at 25°C and
1.00 atm.
a. If the total sample volume was 0.480 L, how many moles of O2 were collected?

Solution :- at 25 C the vapor pressure of the water = 23.8 mmHg * 1 atm /760 mmHg = 0.031316 atm

So the pressure of the O2 = 1 atm – 0.031316 atm = 0.968684 atm

T= 25 C +273 = 298 K

Volume = 0.480 L

Moles of O2 = ?

Using the ideal gas equation we can calculate the moles of the O2

PV= nRT

PV/RT= n

0.968684 atm * 0.480 L / 0.08206 L tamper mol K * 298 K = n

0.019014 mol = n

So the moles of the O2 collected = 0.019014 mol
b. If the same volume of oxygen is collected over ethanol instead of water, does it contain the same number of moles of O2?
Solution :- the moles of O2 collected over ethanol would not be same because the vapor pressure of the different liquids are different. So the moles of O2 would not be same.

6.104. Water and ethanol were removed from the O2 samples in Problem 6.103.
a. What is the volume of the dry O2 gas sample at 25°C and 1.00 atm?
Solution :- moles of O2 = 0.019014 mol

T = 25 C + 273 = 298 K , P = 1 atm

V= nRT/P

V= 0.019014 mol * 0.08206 L atm per mol K *298 K / 1 atm

V= 0.465 L

So the volume of the dry O2 = 0.465 L

b. What is the volume of the dry O2 gas sample at 25°C and 1.00 atm if P (ethanol) = 50 mmHg at 25°C?

Solution :- pressure of O2 = (1 atm – (50 mmHg *1 atm / 760 mmHg) ) = 0.934211 atm

Volume of the dry O2

V= nRT / P

= 0.019014 mol * 0.08206 L atm per mol K * 298 K / 0.934211 atm

= 0.498 L

So the volume of the dry O2 = 0.498 L

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