Hello! This problem is driving me nuts! The vapor pressure of ethanol (C2H5OH) a

Question

Hello! This problem is driving me nuts! The vapor pressure of ethanol (C2H5OH) at 20*C is 44 mmHg, andthe vapor pressure of methanol (CH3OH) at the same temperature is94 mmHg. A mixture of 30.0g of methanol and 45.0 g of ethanolis prepared (assume it behaves as an ideal solution). Calculate the mole fraction of both ethanol and methanol in thevapor above this solution at 20*C. Been stuck on it for an hour, no luck. Any help would begreatly appreciated! Thanks Hello! This problem is driving me nuts! The vapor pressure of ethanol (C2H5OH) at 20*C is 44 mmHg, andthe vapor pressure of methanol (CH3OH) at the same temperature is94 mmHg. A mixture of 30.0g of methanol and 45.0 g of ethanolis prepared (assume it behaves as an ideal solution). Calculate the mole fraction of both ethanol and methanol in thevapor above this solution at 20*C. Been stuck on it for an hour, no luck. Any help would begreatly appreciated! Thanks

Explanation / Answer

First, convert the pressures from mmHg to atm P(EtOH)=0.0579 atm P(MeOH)=.124 atm R=0.08206 L atm/ mol K T=293.15 K L=1 atm (I just chose this arbitarily because of theease) n=PV/RT n(EtOH)=2.41 mmol n(MeOH)=5.14 mmol Mole Fraction= n(molecule)/n(total) Frac(EtOH)=0.319 Frac(MeOH)=0.618

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