Question #8: A 1.00 L flask is filled with 1.45 g of argon at 25 ?C. A sample of

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Question #8: 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 1.050 atm.

Question #9 Part B: A gaseous mixture of O2 and N2 contains 36.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 305 mmHg ? Express you answer numerically in millimeters of mercury.

Question #13 Part A: A syringe containing 1.71 mL of oxygen gas is cooled from 92.1 ?C to 0.5 ?C. What is the final volume Vf of oxygen gas?

Part A What is the partial pressure of argon, Par, in the flask? Express your answer to three significant figures and include the appropriate units. PArValue Units Submit Hints My AnswersGive Up Review Part Part B What is the partial pressure of ethane, Pethane, in the flask? Express your answer to three significant figures and include the appropriate units. PethaneValueUnits

Explanation / Answer

using ideak gas equation

PV=nRT ;

P =nRT/V; = (1.45/40) X 0.0821 X 298

P = 0.887 atm

hence partial pressure of Ar = 0.887 atm

and that of ethane = 1.05 - 0.887 = 0.163 atm

9)

36.8 % by mass = 36.8 g of N2 in 100 g of mixture ......hence it has 63.2 g of O2

or moles of N2 = 36.8 / 28 = 1.314

moles of O2 = 63.2 / 32 = 1.975

hence mole fraction of O2 = moles of O2/ (moles of O2 + moles of N2)

= 1.975 / (1.975 + 1.314) = 0.6

hence mole fraction of N2 = 1-0.6 = 0.4

acc to daltons law of partial pressures

PO2 = XO2 x Ptotal

= 0.6 X 305

= 183 mmHg

13)

assuming that the O2 behaves as ideal gas and pressure remains same inside the syringe ( by expecting that there is no leaking of O2)

hence

PV=nRT

or

PV1 = nRT1 ......(1)

PV2 = nRT2.......(2)

dividing (2)/(1)

V2/V1 = T2/T1

V2/1.71 = 273.5/365.1

V2 = 1.28 ml

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