Question

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d=1.114 g/mL; M= 62.07 g/mol) and water (d=1.00 g/mL) at 20°C. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol as (a) volume percent I % v/v (b) mass percent % w/w (c) molarity м (d) molality (e) mole fraction

Answer 1

Ginen, density of ethylene glycol = 1.lleglon Molar mass = 62.07 ginal denvily of water = 1.00gım density of solution = 1.070g/m2

Firstly, finding the mass of each component so ansurning as the equal volumes mined Mass of water - density x volume cuolent) = 19 x lmx mi Mass of ethylene glycal solute) = 1.114g & Loml mŁ = 1.1149 now, Mass of solution = mass of solute & solent 1971.114g = 2.114 g Now, Volume of solution = mass of solution densiti o solution = 2.1149 -1976m2 1.070gimi

(a) volume Percent = volume of solute , volume of solution E Doo - 1 mLXIO0%. 1.976 mL - 50.6% VIV how (b) finding (b) Mass peruent = mass of solute , 100 1. Mass of solution = 101149 - X 100 2.114 = 52.6% w/w

(1) rolarity = Moles of solute volume of sol.in Moles of solute (met n ) = ginen Mass • Cethylene glyul) rolar Mass - 1.1147 1. 62.07 glmal = 0.017 94 mol : Molarity = 0.01794 x 10ool 1.976 m2 E 9.0871 (d) molality - Moles of solute . Kg solient = 0.01799 x 1000kg = 17,94 m.

(e) Hole fraution 1 = moles of ethylene glycol - Total no. of moles Moles of ethylene glycol = 0.01794 mol Males of water - - almal 3 g/mol -0.055 mol Total no. of moles: 0.01794 +0.055 = 0.07349 i Mole fraition = 0.01794 0.07349 = 0.244