Sample problem 57 calculating gas density a at stp or

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Sample Problem 5.7Calculating Gas DensityPLAN:We can use the molar mass of CO2to find its density from the ideal gas equation. PROBLEM:Find the density (in g/L) of CO2 (g) and the number of molecules per liter(a)at STP and (b)at room conditions (20.°C and 1.00 atm).
SOLUTION:
P
5-36Sample Problem 5.7(b) At 20.°C and 1.00 atm:SOLUTION:Mx PRTd==44.01 g/molx 1.00 atm0.0821 x 293 Kmol·Katm·L= 1.83 g/L1.83 g CO21 L1 mol CO244.01 g CO2xx6.022 x 1023molecules1 mol= 2.50 x 1022molecules CO2/LT= 20.°C + 273.15 = 293 K
Sample Problem 5.8Finding the Molar Mass of a Volatile LiquidPROBLEM:An organic chemist isolates a colorless liquid from a petroleum sample. She places the liquid in a preweighed flask and puts the flask in boiling water, causing the liquid to vaporize and fill the flask with gas. She closes the flask and reweighs it. She obtains the following data:Calculate the molar mass of the liquid.Volume (V) of flask = 213 mLT= 100.0°CP= 754 torrmass of flask + gas = 78.416 gmass of flask = 77.834 g
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5-39Sample Problem 5.8SOLUTION:mof gas = (78.416 - 77.834) = 0.582 g M= mRTPV=0.582 g xatm·Lmol·K0.0821x 373 K0.213 Lx0.992 atm= 84.4 g/mol1 L103mLV= 213 mL x= 0.213 LT= 100.0°C + 273.15 = 373.2 K1 atm760 torrP= 754 torr x= 0.992 atm
5-40Gases mix homogeneously in any proportions.Each gas in a mixture behaves as if it were the only gas present.The pressure exerted by each gas in a mixture is called its partial pressure.Dalton’s Law of partial pressures states that the total pressure in a mixture is the sum of the partial pressures of the component gases.The partial pressure of a gas is proportional to its mole fraction:PA= XAx PtotalXA=nAntotalMixtures of Gases
5-41Sample Problem 5.9Applying Dalton’s Law of Partial PressuresPROBLEM:In a study of O2uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mole % N2, 17 mole % 16O2, and 4.0 mole % 18O2. (The isotope 18O will be measured to determine the O2uptake.) The pressure of the mixture is 0.75 atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2in the mixture.PLAN:Find Xand Pfrom Ptotaland mol % 18O2. 18O218O2divide by 100multiply by Ptotalpartial pressure P 18O2mole % 18O2mole fraction, X18O2
5-42Sample Problem 5.9SOLUTION:= 0.030 atmP= Xx Ptotal= 0.040 x 0.75 atm 18O218O2= 0.040X18O2=4.0 mol % 18O2100
5-43T(0C)T(0C)0510121416182022242628303540455055606570758085909510055.371.992.5118.0149.4187.5233.7289.1355.1433.6525.8633.9760.04.66.59.210.512.013.615.517.519.822.425.228.331.842.2Table 5.2 Vapor Pressure of Water (P) + at Different TH2OP (torr)H2OP (torr)H2O
5-44Figure 5.12Collecting a water-insoluble gaseous product and determining its pressure.

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