Question

If 0.720 molmol  of a nonvolatile nonelectrolyte are dissolved in 3.70 molmol of water, what is the vapor pressure PH2OPH2O of the resulting solution? The vapor pressure of pure water is 23.8 torrtorr at 25 ∘C∘C .

Express your answer with the appropriate units

A solution is composed of 1.60 molmol cyclohexane (P∘cy=97.6 torrPcy∘=97.6 torr) and 2.90 molmol acetone (P∘ac=229.5 torrPac∘=229.5 torr). What is the total vapor pressure PtotalPtotal above this solution?

Express your answer with the appropriate units

Answer 1

a schetton a) Given Data Moles of solute = 0720 mo/ Moles of solvent (426) = 3.70 mal vapoure pressure of pure waters Priso = 23-8 torr. Mole fraction of H, 0 (XH20) Moles of H2O Moles of solute + Moles of H₂O TH2o 3.70 mol or 720 mol +370 mol TH2O=0.8371 According to Rowl's Law where pr_pox po- vapor pressure of Pune Compound. p= vapour pressure of compound in solution sea mole fraction.

Prizo = P20 X Xrizo = 23.8tors x0.8371 So, = 19.923 torr. The rapor pressure of water in solution is Przo = 19.923 torr. Data 1-60 mol 2-90 mol b) Given Moles of cyclohexone Moles of acetone = vapor pressure of pure cyclohexane Pey=97-6tors rapor pressure of pure acetone P&c=229.5 ton Protal + Pac Sey= Paccyx Xey Pac Paco Xac Pay رن زنة) Prestal - Pey x Key + Pac fac Xey: Moles of cyclohexane of cyclohexane clohexane trdes of acetone Moles 1.60 mol 0.355 m (2.904 1:60) moi

Xac = Moles of acetone 4 Motos of cyclohetone moles of acetone Xac 2.90 mol (1.60+2096Jmoi 0.64 4 4 nol Protal= (97.6 torr X 6-3593+ (229.5x 6.699 = 34.70 tors + 147.89 torr • 182.60 torr PTotal = 182.6 torr.