I need help solving this question.. An aqueous solution of ethylene glycol (C2H6O2) IS 40% ethylene glycol by mass, and has a density of 1.05g/ml. what are the molarity, molality fraction of the solution?
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A gas occupies 10 liters at standard conditions. what volume will it occupy at 20 C and 700 torr?
I don't know what you mean by molality fraction.
mass of 1,000 mL = 1.05 g/mL x 1000 = 1050 gram. That is 40% ethylene glycol; therefore,
1050 x 0.40 = 420 g ethylene glycol and 420 g is 420/molar mass ethylene glycol. That will give you moles/L which is M.
To find molality,
1050 grams soln - mass ethylene glycol (420 g) = 630 g water.
molality = moles/kg soln. Convert 420 to moles ethylene glycol and divide by 0.630 = ?? molal
mass of 1,000 mL = 1.05 g/mL x 1000 = 1050 gram. That is 40% ethylene glycol; therefore,
1050 x 0.40 = 420 g ethylene glycol and 420 g is 420/molar mass ethylene glycol. That will give you moles/L which is M.
To find molality,
1050 grams soln - mass ethylene glycol (420 g) = 630 g water.
molality = moles/kg soln. Convert 420 to moles ethylene glycol and divide by 0.630 = ?? molal
For the gas problem use PV = nRT.
2 answers