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

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:

(1)

The moles of ethanol are:

18,00mL×× = 0,3083 mol Ethanol.

Moles of benzoic acid:

12,55 g× = 0,1028 mol benzoic acid.

Thus, mole fraction of solvent, X, is:

= 0,7499

Replacing this value in (1):

= 10,27 kPa

I hope it helps!

2b2t

explanation:

2b2t