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Answer: The vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
and
where, x = mole fraction in solution
= pressure in the pure state
According to Dalton's law, the total pressure is the sum of individual pressures.
moles of ethanol=
moles of methanol=
Total moles = moles of ethanol + moles of methanol = 1.7 +3.0 = 4.7
Thus the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.