Question

Part A What is the vapor pressure in mmHg of a solution prepared by dissolving 5.00 g of benzoic acid (CzH6O2) in 100.00 g of ethyl alcohol (C2H60) at 35 °C? The vapor pressure of pure ethyl alcohol at 35 °C is 100.5 mmHg. Assume that benzoic acid does not ionize in this solvent. V ACO R O P ? Psoln = mm Hg Submit Request Answer Provide Feedback Next >

Answer 1

Molaro mass of benzoic acid = 122.12 g/mol Number of motes of benzoic acid (ha) = 5.009 122.12 g/mol = 0.041 mol Molaro mass of ethyl alcohol = 46.07 g/mol Number of moles of ethyl alcohol th - 100.00 g. 46.07 g/mol Total numbers of moles (ht) = hitha = 2.170 mol = (2.170 +0.041) mol = 2.211 mol Mole fraction of Solvent, ethyl alcohol (x solvent) = - Hero VW nT - 2.170 mol 2. 211 mol vapor proessure of O = 0.981 I Solvent solution (Pson) = X solvent & P solven = 0.981 x 100.5 mm Hg = [98.59 mm Hg