Question 22 A 1.00 L flask is filled with 1.45 g of argon at 25 °C. A sample of

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Question 22 A 1.00 L flask is filled with 1.45 g of argon at 25 °C. A sample of ethane vapor is added to the same flask until the total pressure is 1.350 atm Part A What is the partial pressure of argon, P in the flask? Ar, Express your answer to three significant figures and include the appropriate units Par Value atm Submit Hints My Answers Give U Review Part incorrect; Try Again; 3 attempts remaining Part B What is the partial pressure of ethane, P thane, in the flask? Express your answer to three significant figures and include the appropriate units Pethane Value atm Submit Hints My Answers Give Up Review Part incorrect; Try Again; 3 attempts remaining

Explanation / Answer

The partial pressure of a gas in a mixture of gases is the pressure exerted by the gas if alone occupies the entire volume available to it.

Part A

CALCULATION OF THE PARTIAL PRESSURE OF ARGON Ar:

pV = nRT

p = nRT / V

n = number of moles of Ar

   = Mass of Ar in gram / Atomic mass of Ar in gram

   = 1.45 g / 39,9 g/mol = 0.0363 mol

R = 0,0821 L atmos/Kmol

T = 250 C = (273+25) K = 298 K

V = 1.00 L

p =0.0363 mol x 0,0821 L atmos/Kmol x 298 K / 1.00 L = 0.888 atmos

The partial pressure of Ar = 0.888 atmos

pAr = 0.888 atmos

Part B

CALCULATION OF THE PARTIAL PRESSURE OF ETHANE:

According to the Dalton's law of partial pressues, The total pressure of a mixture of gases is equal tothe sum of the

partial pressures of the constituent gases.

Partial pressure of Ar + Partial pressure of Ethane = Total Pressure

Partial pressure of Ethane = Total Pressure - Partial pressure of Ar

Total Pressure = 1.350 atmos

Partial pressure of Ar = 0.888 atmos

Partial pressure of Ethane = 1.350 atmos- 0.888 atmos = 0.462atm

pethane = 0.462atm

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