Question

The density of an aqueous solution containing 10.0 percent of ethanol (C_2H_3OH) by mass is 0.984 g/mL. (a) Calculate the modality of this solution. (b) Calculate its modality. (c) What volume of the solution would contain 0.125 mole of ethanol?

how do i do part c?

Answer 1

a) Mass per cent of C2H.OH solution-10 % 10% (m/m) solution means 10 g of C2H5OH solute is present per 100 g of solution Mass of C,H,OH solute -10 g 1 mol 「mỏI C2H5 Noof moles of C2H5OH= 10 g× = 0.217 mol C2H5OH Mass of solution -100 g Mass of solute -10 g Mass of water solvent -m ass of soluti on -mass of solute -100 g-10g -90 g-0.090 kg no of moles of solute mass of solvent in kg 0.217 mol = 2.41 mol/kg-2.41m 0.090 kg Molality of solution = b) No of moles of C2H5OH solute = 0.217 mol C2H5OH Mass of solution -100 g Denisty of solution -0.984 g/mL Volume of solution= 100 g × 1 mL-102 mL×1000mL 1 L 0.984 g = 0.102 L 0.217 mol 0.102 L Molar concentrati on - = 2.13 mol/L = 2.13 M c) molarity of C2H,OH solution 2.13 M 2.13 mol/L No of moles of C,H,OH solute 0.125 mol Volume of solution = 0.125 mol C2H5OH × 1 L 2.13 mol = 0.0587 L x 1000 mL 1 L 58.7 mL