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

Answer:

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

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×$\frac{0,789g}{1mL}$×$\frac{1 mol}{46,07g}$ = 0,3083 mol Ethanol.

Moles of benzoic acid:

12,55 g×$\frac{1mol}{122,12g}$ = 0,1028 mol benzoic acid.

Thus, mole fraction of solvent, X, is:

$\frac{0,3083 mol}{0,3083mol+0,1028mol}$ = 0,7499

Replacing this value in (1):

$P_{solution} = 0,7499*13,693kPa$ = 10,27 kPa

I hope it helps!