A gas sample is known to be a mixture of ethane and butane. A bulb having a 230.
ID: 733438 • Letter: A
Question
A gas sample is known to be a mixture of ethane and butane. A bulb having a 230.0-cm3 capacity is filled with the gas to a pressure of 97.5Explanation / Answer
PV = nRT =>97.5*1000*230*10^-6 = n*8.314*(23.1+273.15) =>n = 9.104667*10^-3 m/M = 9.104667*10^-3 =>m = 9.104667*10^-3*M =>0.3554 = 9.104667*10^-3*M =>M = 0.3554/9.104667*10^-3 = 39.03492515.......(1) Let mass of ethane be x g and butane be y g Then x+y = 0.3554........(2) M = (30.07*x) + (58.12*y)/(x+y) =>39.03492515x + 39.03492515y = 30.07x + 58.12y =>8.964925151x = 19.08507485y =>x = 2.128860479y........(3) Substitute in (2) to get, 3.128860479y = 0.3554 =>y = 0.1135876791 g and x = 0.2418123209 Percent of butane in mixture = (y/(x+y))*100 = 31.96%
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