The vapor pressure of ethanol (C2H5OH) at 20°C is 44 mmHg, and the vapor pressure of methanol (CH3OH) at the same temperature is 94 mmHg. A mixture of 26.9 g of methanol and 47.3 g of ethanol is prepared and can be assumed to behave as an ideal solution. Calculate the vapor pressure of methanol and ethanol above this solution at 20°C. Be sure to report your answers to the correct number of significant figures.

See explanation below

Explanation:

We have a mixture of two alcohols here. Methanol and Ethanol. In order to know the partial pressures of these gases in the mixture, we first need to calculate the number of moles of both alcohols by separate, using the molar mass of Methanol and Ethanol.

Let MEthanol be A and Ethanol B.

MMA = 32.04 g/mol

MMB = 46.07 g/mol

nA = 26.9 / 32.04 = 0.8396 moles

nB = 47.3 / 46.07 = 1.0267 moles

With these moles, we can calculate the molar fraction of each, The molar fraction is calculated with the expression:

Xa = nA / nT

and nT is the sum of the moles of methanol and ethanol, so:

nT = 0.8396 + 1.0267 = 1.8663

The molar fraction of each alcohol is:

Xa = 0.8396 / 1.8663 = 0.4499

Xb = 1.0267 / 1.8663 = 0.5501

Now that we have the molar fraction, we can calculate the partial pressure of each alcohol and then, of the whole solution:

P = X * Pp

so, the pressure for methanol and ethanol are:

PA = 0.4499 * 94 = 42.2906 mmHg

PB = 0.5501 * 44 = 24.2044 mmHg

So the total pressure of the solution is:

P = 42.2906 + 24.2044 = 66.4950 mmHg