At 293 K, methanol has a vapor pressure of 97.7 Torr and ethanol has a vapor pressure of 44.6 Torr. What would be the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K?

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.

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