Question

5. The total volume of an ethanol solution at 25 °C containing 1.000 kg of water is experimentally found to obey the following expression V/ml = 1002.9 + 54.675 -0.3639b2 + 0.02825b3, Where b is the numerical value of the molality of ethanol in the solution. At what molality would the partial molar volume of ethanol be the minimum? 1. Cannot be answered, as not enough information is given. ANS D 2. b = 0 mol/kg 3. b = 21.74 mol/kg (approx.) 4. b = 4.29 mol/kg (approx)

Please provide explanation

Answer 1

" Vm = 100209 + 54.676 -0.36396² +0.02825b3 - Ethanol where; b molality of . in solution . 0 1 To find molality, where partial molar volume of Ethanol be the minium; we will diffrentiate. ero worot to 'b' u dvimo = d $100269 + 54-67b-o. +0.02825b3Y 92 db - dumi - ot 54.67 (1 - 2X003639b db + 0.02825 (36²) E) 00 08475b²_007278b + 54067 duml - db for vms to be minimum dumi - o - 0.084756²007278b +54.67 dh

0.084756² -0.7 278 b + 54.67=0 By solving quadratic equation we will get b= 402938 + 25.03271 (i complex root - let's take real value of complex soot - b= 40 2938 mollkg At this vmi minimum