if you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of codeine?

Question

Maddox Hines

Chemistry

27 May, 23:22

if you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of codeine?

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Answers (1)

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Julian Bauer · Answer 1

The mole fraction of codeine is 5.4%

Explanation:

Mole fraction = Mole of solute / Total moles (Mole solute + Mole solvent)

Solute (Codeine)

Molar mass 299.36 g/m

Mass / Molar mass = Mole → 46.85 g / 299.36 g/m = 0.156 moles

Solvent (Ethanol)

Molar mass 46.07 g/m

125.5 g / 46.07 g/m = 2.724 moles

Mole fraction (Codeine) 0.156 / (0.156 + 2.724) → 0.054