1 atm 7095 mmhg 760 mmhg 0933552631 atm pv nrt

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1 atm709.5 mmHg760 mmHg= 0.933552631 atm PV = nRT Solving for V: V= L•atm0.0605100 mol0.0821291 Kmol•K= 0.933552631 atmnRTP= 1.5485 = 1.5 L H25-16
5.53 Plan:Rearrange the ideal gas law to calculate the density of the air from its molar mass. Temperature must be converted to kelvins and pressure to atmospheres. Solution:P= 744 torr T= 17°C + 273 = 290 K or T= 60°C + 273 = 333 K Mof air= 28.8 g/mol d= unknown Converting Pfrom torr to atm: P= 1 atm744 torr760 torr= 0.978947368 atm PV = nRT Rearranging to solve for density: At 17°C 0.978947368 atm28.8 g/molL•atm0.0821290 Kmol•KPd =RTM= 1.18416 = 1.18 g/LAt 60.0°C 0.978947368 atm28.8 g/molL•atm0.0821(333 K)mol•KPd =RTM= 1.03125 = 1.03 g/L5.54 Plan:Solve the ideal gas law for molar volume, n/V. Pressure must be converted to atm and temperature to K. Solution:P= 650. torr T= –25°C + 273 = 248 K n/V= unknown Converting Pfrom torr to atm: P= 1 atm650. torr760 torr= 0.855263157 atm PV = nRT Solving for n/V: 0.855263157 atmL•atm0.0821248 Kmol•KnP=VRT= 0.042005 = 0.0420 mol/L5.55 Plan:The problem gives the mass, volume, temperature, and pressure of a gas; rearrange the formula PV= (m/M)RTto solve for the molar mass of the gas. Temperature must be in Kelvin and pressure in atm. The problem also states that the gas is a hydrocarbon, which by, definition, contains only carbon and hydrogen atoms. We are also told that each molecule of the gas contains five carbon atoms so we can use this information and the calculated molar mass to find out how many hydrogen atoms are present and the formula of the compound. Solution:V= 0.204 L T= 101°C + 273 = 374 K P= 767 torr m= 0.482 g M= unknown Converting Pfrom torr to atm: P= 1 atm767 torr760 torr= 1.009210526 atm = mPVRTMSolving for molar mass, M:L•atm0.482 g0.0821 374 Kmol•K= 1.009210526 atm0.204 LmRTPVM= 71.8869 g/mol (unrounded) 5-17
The mass of the five carbon atoms accounts for [5(12 g/mol)] = 60 g/mol; thus, the hydrogen atoms must make up the difference (72 – 60) = 12 g/mol. A value of 12 g/mol corresponds to 12 H atoms. (Since fractional atoms are not possible, rounding is acceptable.) Therefore, the molecular formula is C5H12. 5.56 Plan:Solve the ideal gas law for moles of air. Temperature must be in units of kelvins. Use Avogadro’s number to convert moles of air to molecules of air. The percent composition can be used to find the number of molecules (or atoms) of each gas in that total number of molecules. Solution: V= 1.00 L T= 25°C + 273 = 298 K P= 1.00 atm n= unknown PV = nRT Solving for n: Moles of air =n =RTPV1.00 atm1.00LL•atm0.0821298 Kmol•K= 0.040873382 molConverting moles of air to molecules of air: Molecules of air = 236.022x10molecules0.040873382 mol1 mol= 2.461395x1022molecules Molecules of N2= 222 78.08% Nmolecules2.461395x10air molecules100% air= 1.921857x1022= 1.92x1022molecules N2Molecules of O2= 222

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