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

** the answer is 10.27 but I'm not sure how to get there. Please show work!!

Answer 1

ethanol mass = vol x density = 18 ml x 0.789 g/ml = 14.202 g

ethanol moles   = mass/Molar mass of ethanol = 14.202 g/ 46 g/mol = 0.30874

benzoic acid moles = mass / molar mass = 12.55 g / 122.12 g/mol = 0.10277

mol fraction of benzoic acid = ( benzoic moles) / ( ethanol moles + benzoic acid moles)

                  = ( 0.10277 / 0.10277 + 0.30874) = 0.25

By raults law

( Po-Ps) = Po x Xb    where Po = Vapor pressure of pure ethanol = 13.693 kpa

Xb = mol fraction of benzoic acid = 0.25 , Ps = vapor pressure of solution

13.693 - Ps = 13.693 x 0.25

Ps = 10.27 kpa    is vapor pressure of solution having benzoic acid in ethanol