Calculate the molarity, molality and mole fraction of ethanol in its 20% aqueous solution. (No other values are given).
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X(EtOH) = 0.0891

X(H₂O)   = 0.9109

Explanation:

Given 20% aqueous solution ethanol.

Ethanol => C₂H₅OH (or, EtOH) => molecular weight = 46 g//mole

20% solution by weight = 20 grams EtOH / 100 grams of solution

20 grams EtOH / 100 grams of solution = 20 grams EtOH + 80 grams H₂O

moles EtOH = (20g/46g/mol) = 0.4348 mole

moles H₂O  = (80g/18g/mol) = 4.4444 mole

moles EtOH + moles H₂O = 0.4348 + 4.4444 = 4.8792

X(EtOH) = 0.4348/4.8792 = 0.0891

X(H₂O)   = 4.4444/4.8792 = 0.9109

∑mole fractions must equal 1.000 => 0.0891 + 0.9109 = 1.000 (check!)

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