Question

What is the vapor pressure (in kPa) of ethanol, CH3CH2OH, over a solution which is composed of 18.00 mL of ethanol and 12.55 g of benzoic acid, C6H5COOH, at 35ºC ?

Enter your number with two digits past the decimal.

•Pºethanol at 35ºC = 13.693 kPa

•Density of ethanol = 0.789 g/mol, Molar mass of ethanol = 46.07

•Molar mass of benzoic acid = 122.12 g/mol

Answer 1

The vapor pressure of ethanol in the solution is 10,27 kPa

Step-by-step explanation:

To obtain the vapor pressure of a solution it is necessary to use Raoult's law:

`P_(solution) = X{solvent}P_(0solvent)` (1)

The moles of ethanol are:

18,00mL×
`(0,789g)/(1mL)`×
`(1 mol)/(46,07g)` = 0,3083 mol Ethanol.

Moles of benzoic acid:

12,55 g×
`(1mol)/(122,12g)` = 0,1028 mol benzoic acid.

Thus, mole fraction of solvent, X, is:

`(0,3083 mol)/(0,3083mol+0,1028mol)` = 0,7499

Replacing this value in (1):

`P_(solution) = 0,7499*13,693kPa` = 10,27 kPa

I hope it helps!