Question

You are asked to prepare an aqueous solution of ethylene glycol (HOCH2CH2OH) with a mole fraction of 0.192.

a) If you use 645 g of water, what mass (in g) of ethylene glycol should you use?

b) What is the molality of the resulting solution?

Answer 1

mole fraction of ethylene glycol     = 0.192

   mole fraction of ethylene glycol    = X eth   = n eth/n eth + n H2O

a. no of moles of H2O (n H2O)     = W/G.M.Wt

                                                        = 645/18   = 35.84 moles

X eth   = n eth/n eth + n H2O

0.192   = n eth/ n eth + 35.84

0.192*( n eth+35.84)   = n eth

n eth     = 8.52

no of moles of ethylene glycol ( n eth) = 8.52 moles

mass of ethylene glycol      = no of moles * gram molar mass

                                           = 8.52*62   = 528.24g

b. molality   = no of moles of solute *1000/weight of solvent in g

                  = 8.52*1000/645   = 13.2m >>>>answer