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

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?

Expert Solution

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

queen_honey_blossom answered 2 years ago

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