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what is the molality of an antifreeze solution containing 15.4 grams ethylene glycol in 84.6 grams of water?
 

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Kaiusviz
Molality = mol of solute / kg of solvent Mol of solute = 15.4 g C2H6O2 × ( 1 mol / 62 g of C2H6O2) Mol of solute = 0.248 mol Kg of solvent = 84.6 g of H2O or 0.0846 kg of H2O Molality = (0.248 mol) / (0.0846 kg) = 2.931 molal