CHAPTER 10: ANSWERS TO ASSIGNED PROBLEMS

1 1 CHAPTER 10: ANSWERS TO ASSIGNED PROBLEMS Hauser- General chemistry I revised 8/03/08 The typical atmospheric pressure on top of Mt. Everest (29028 ft) is about 265 torr. Convert this pressure to (a) atm 265 torr ( 1 atm / 760 torr) = atm (b) mm Hg 265 torr ( 760 mm Hg / 760 torr) = 265 mm Hg A scuba divers tank contains kg of O2 compressed into a volume of L. (a) Calculate the gas pressure inside the tank at 9 C. PV = nRT so P = nRT / V convert kg to 290 g O2 convert 9 C to K [9 C + = K ] (do not pre-round) 290 g O2 ( 1 mol O2 / g ) ( L atm / mol K) ( K) P = ---------------------------------------- ---------------------------------------- ----- = 91 atm L ( 2 SF) Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters.

2 Suppose that the volume of particular sample of Cl2 gas is L at 895 torr and 24 C. (a) How many grams of Cl2 are in the sample? Need to solve for moles to get grams. PV = nRT so n = PV / RT convert 24 C to K [24 C + = K ] (do not pre-round) convert 895 torr to atm since R demands it [ (895 torr) ( 1 atm / 760 torr) ] ( L) n = ---------------------------------------- ----------- = mol Cl2 ( L atm / mol K) ( K) mol Cl2 ( g / 1 mol Cl2 ) = g = g Cl2 2 ADDITIONAL EXERCISE #10 An ideal gas originally at atm and 66 C was allowed to expand until its final conditions were a volume of 94 mL, pressure of atm, and temperature of 45 C.

3 What was the initial volume of the gas? T1 = 66 C + = K T2 = 45 C + = K P1V1 / T1 = P2 V2 / T2 Solve for V1 V1 = P2V2 T1 / T2 P1 ( atm) (94 mL) ( K) V1 = --------------------------------------- = mL = 71 mL ( K) ( atm) Which gas is most dense at atm and 298 K? CO2, N2O, or Cl2? Explain. GAS DENSITY IS DIRECTLY RELATED TO MOLECULAR WEIGHT. NO CALC. NECESSARY SINCE CONDITIONS ARE CONSTANT. MOST DENSE GAS OF THESE THREE IS Cl2. Cl2 = g/mol N2O = g/ mol CO2 = g/mol (a) Calculate the density of NO2 gas at atm and 35 C. Convert temp: 35 C + = K (really 3 SF) USE: d = Mm P / R T Mm NO2 = g / mol d = ( g / mol )( atm ) / ( L atm / mol K) ( K) = = g/L (3 SF) (b) Calculate the molar mass of a gas if g occupies L at 685 torr and 35 C.

4 D = Mm P / R T so Mm = d R T / P ( g / L) ( L atm/ K mol) ( K) Mm = ---------------------------------------- -------------------------- = = g/mol ( 685 torr ) ( 1 atm / 760 torr) 3 The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas: C6H12O6 (aq) + O2 (g) 6 CO2 (g) + 6 H2O (l) Calculate the volume of dry CO2 produced at body temperature (37 C) and atm when g of glucose is consumed in this reaction. PV = nRT so V = nRT / P Need moles of glucose converted to moles of the product gas (CO2). USE STOICHIOMETRY: molecular weight calc: C 6 X = H X 12 X = O 6 X = sum = g C6H12O6 ( 1 mol C6H12O6 / g) ( 6 mol CO2 / 1 mol C6H12O6) = mol CO2 gas.

5 Convert temp: 37 C + = K ( mol) ( L atm / mol K) ( K) V = ---------------------------------------- --------------------- = = L CO2 atm ADDITIONAL EXERCISE #11 Based on the following balanced equation, what volume of oxygen gas would be expected from complete reaction of grams of KCLO3 at STP ? 2 KCLO3 (s) 2 KCL (s) + 3 O2 (g) g KClO3 (1 mol KClO3 / g KClO3) ( 3 mol O2 / 2 mol KClO3) ( L / 1 mol O2 gas at STP) = = L O2 gas (1 MOLE OF ANY GAS @ STP HAS L VOLUME.) A mixture of gases contains mol N2, mol O2, and mol CO2. If the total pressure of the mixture is atm, what is the partial pressure of each component?

6 DALTON'S LAW OF PARTIAL PRESSURES: P total = P gas 1 + P gas 2 + P .. APPROACH: Sum the moles and find the percentage due to each gas. mol N2 / mol (X 100%) = of atm = = atm N2 mol O2 / mol (X 100%) = of atm = = atm O2 mol CO2 / mol (X 100%) = of atm = = atm CO2 [sum = mol] [P total = ~ atm] 4 Hydrogen has two naturally occurring isotopes, 1H and 2H. Chlorine also has two naturally occurring isotopes, 35Cl and 37Cl. Thus, hydrogen chloride gas consists of four distinct types of molecules: 1H35Cl, 1H37Cl, 2H35Cl, and 2H37Cl. Place these four molecules in order of increasing rate of effusion. THE RATE OF EFFUSION IS INDIRECTLY RELATED TO MOLECULAR WEIGHT.

7 THE LIGHTER THE PARTICLE, THE FASTER IT EFFUSES. slowest to fastest 2H37Cl < 1H37Cl < 2H35Cl < 1H35Cl ADDITIONAL EXERCISE #12 Compare the relative rate of effusion of a molecule of oxygen and a molecule of xenon. Which gas travels faster? How much faster? GRAHAM'S LAW OF EFFUSION rate gas 1 MW gas 2 ------------ = square root of ------------- rate gas 2 MW gas 1 rate of O2 MW Xe g/mol ------------ = square root of ------------ = ---------------- = square root of = rate of Xe MW O2 g /mol ANSWER = = INTERPRET: THE OXYGEN EFFUSES TIMES FASTER THAN XENON. MAKES SENSE: THE LIGHTER GAS ALWAYS TRAVELS FASTER. Notes: Use O2, not O.

8 Reported 4 SF based on molecular weights here, but no obvious SF rule to follow here since all "exact" numbers. (a) List two experimental conditions under which gases deviate from ideal behavior. VERY HIGH PRESSURE and/or VERY LOW TEMPERATURE (b) List two reasons why the gases deviate from ideal behavior. DIFFERENT GAS MOLECULES HAVE DIFFERENT VOLUMES AND DIFFERENT ATTRACTIVE FORCES. THESE COULD CAUSE NON-IDEAL BEHAVIOR UNDER THE EXTREME CONDITIONS LISTED ABOVE.