In a 61.0 -g aqueous solution of methanol, CH 4 O , the mole fraction of methanol is 0.130. What is the mass of each component? mass of CH 4 O : g CH 4 O mass of water:

Answer 1

I am using y for mole fraction and x for mass fractions. Components of your system: MeOH (methanol) and H2O. MM = molar mass. MM MeOH = 32 g/mol, MM H2O = 18 g/mol. Sum of mole fractions is 1, and sum of mass fractions is also 1. Definition of a mole fraction: mol component/mol total. Definition of a mass fraction: mass component/mass total.

yMeOH=0.130, therefore yH2O=0.870; assuming a 100 mole solution, this means you have 13.0 mol MeOH and 87.0 mol H2O.

13 mol MeOH * 32 g/mol = 416 g MeOH

87 mol H2O * 18 g/mol = 1566 g H2O

Total mass = 1566+416 = 1982 g. Can compute mass fractions from here:

xMeOH = 416gMeOH/1982g total = 0.210

xH2O = 1-xMeOH OR 1566gH2O/1982g total (answers will match) = 0.790

Overall mass is 61.0g. Mass of MeOH = xMeOH * overall mass because xMeOH = xMeOH/overall mass, so overall mass cancels. 12.81 g MeOH. Mass of water will be 61.0g-12.81g OR xH2O *overall mass (same answer), 48.19 g H2O.