Question

Givent that the molecular mass of ethyl alcohol is 46 g/mol and that of water is 18 g/mol, how many grams of wthyl alcohol must be mixed with 100 mL of watre for the mole fraction of ethyl alcohol to be 0.2?

Answer 1

Mole fraction of ethyl alcohol = 0.2

Molecular mass of ethyl alcohol = 46 g/mol

Molecula mass of water = 18 g/mol

Volume of water = 100 ml

Mole fraction of ethyl alcohol = moles of ethyl alcohol/(moles of ethyl alcohol + moles of water)

                                            = (wt/gram molecular weight of C₂H₅OH)/(wt/GMW of C₂H₅OH +wt/GMW of water)

                                       Density of water = 1g/ml

                So, weight = density * volume = 1*100 = 100 g

From this, number of moles of water = 100/18 = 5.55 moles

substituting in the mole fraction equation, 0.2 = (wt/gram molecular weight of C₂H₅OH)/(wt/GMW of C₂H₅OH +5.55)

          0.2 = (wt/46)/[(wt/46) +(5.55)]

0.087*wt of C₂H₅OH = 5.55

        wt of C₂H₅OH = 63.22 g of ethyl alcohol is required