A gas mixture contains 62 g of carbon monoxide and 680 g of carbon dioxide. What is the mass%, molality and mole fraction of carbon monoxide?

Question

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Answer 1

moles carbon monoxide (CO) =62 g x 1 mole/28 g = moles CO

moles carbon dioxide (CO2) = 680 g x 1 mole/44 g = moles CO2

 

mass % CO = 62 g/(62 g + 680 g) X (100% ) = 62 g/742 g (x100%) = 8.36%

mole fraction CO = moles of CO/total number of moles  (x100%): use values determined above

 

molality of CO = moles of CO (as determined above) per kg solvent. In this case, you can probably consider the CO2 to be the solvent as it is present in the greatest amount. Thus, you have moles of CO (as calculated above)/0.680 kg = m