Question

Question 10:  

We have an aqueous solution with a mass percent of 42.7 % for C₂H₅OH (C₂H₅OH is the only solute). The density of this solution is 0.942 g/mL. Calculate the molarity, molality, and mole fraction of C₂H₅OH in this aqueous solution.  

Answer 1

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Solution:

Solution: Basis: 100 gm of aqueous solution. CH₃OH present 42.7% by mans Molecular weight = 12x2 + 5x1 + 16x1 * = 46 gen/mel Since, no. of carbon atom - Molecular wil. of C =12 no. of H atom = 6 Molo wt of H atom=1 & no. of O-atam = 1 molecular wt. of 0 = 16 So in the solution C₂ H₂OH = 42.27 gm & HO=57.3gm. Now, Molarity of moles of solute Volume of the solution in L. Molality no. of moles of solute Amount of solution in Kg. 3.18 mol 42.7 no, of moles of G H OH = 0,93 mol. 46 no. of moles of H₂O = 57.3 18 Amount of solution - 100 gm Given density of the solution - 0.942 gelme.

Total moles in the solution = nets BH' 1 mL 0.942 gm contains so 100 gmi I x 100 = 106.15 mL 0.942 volume of the solution = 106.15mL = 0.106 L. & mans of the solution, = 100gm = Ool kg Molarity mol solute volume of solution in L = 8.77 mol/ 0.106 molality Mole of solute Amount of solution inkg 0.93 = 9.3 m molling + NHO 0.93 +3,18 = 4.11 mol. - nes HS OH + mole fraction of Ç H₂ OH NGH SOH MHO 0.93 = 0.226 4:11