Question

A solution is made by mixing 41.0 mL of ethanol, C2H6O, and 59.0 mL of water. Assuming ideal behavior, what is the vapor pressure of the solution at 20 ∘C? The relevant values at 20 ∘C are included in the table.
Liquid   Density (g/mL)   Vapor Pressure, ?∘ (Torr)
ethanol   0.789   43.9
water   0.998   17.5

Answer 1

* Given: Volume of ethanol = 410 mecy) Volume of water: 59.0 m (V2) density of ethenol = 0.789 g/me density of walde - 0.998 g/mu. Vapour Pressure ethanol = 43.9 Force) Vapour prexure & watel = 17 sturen) * To Find: - Vapour pressure of ethanolt Water =? * calculation - Accoading to Raoult taw, Pi=Pixi and P2 = P2x2. pils pi – Vapour pressure of ethanol of water respectively. 044x2 - mole fraction of o ethanol & water resp. Now, where, mole fraction =OX) - no of moles of ethanol no of moles of ethand & Water. and, no of moles of ethanol =co) - mass of ethanol Votre of ethanol. Mdar mass .. nismo & mass = density x volume mass of ethanol = 0.789 441 mass of ethano) = 32.34-9 g) Moler mass of ethanol = C2H60 = (2x12) +C\x6)+C1X16) M = 24+ 6+16= 46 g Mi=468 0,= = 32:34 46. ni= 0.7032) --- – (moles of ethanol)

no of moles of water = n2 = mass of water Molar mass of watet and mass = Volumex density = 59x 0.998 = 58882. go! mass Molar Mass C H20) = 1X2 +6=18g. 12 = = 58.882 = 3.2712 18 m2 = 3.2712|- - --- moles of water. Now, el = 0 - 0.7032 M2 ni+D2 0.7032+ 3.2712 al = 0.1769 Similarly, (water) = 12 - the 708 3.2712 ni+n2 0.7032+3-2712 202 = 0.8230 Now, Pi= pixel Picethanol) = 43.980.1769 JP = 7.7659 Toor] Similarly, P2 (wates) = P2 xx 2 = 17.5X0.8230 TP2 = 14.4025 PT (ethanaltwater) = Pitp = 7.7659 +14.4025 IPT = 22.1684 Toor]