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!

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.