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Antifreeze has ethylene glycol, C₂H₆O₂ (62 g mol⁻¹), dissolved in water…

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Antifreeze has ethylene glycol, $\text{C}_2\text{H}_6\text{O}_2$ ( $62\ \text{g mol}^{-1}$ ), dissolved in water. A mechanic mixes $62\ \text{g}$ of it with $250\ \text{g}$ of water. Find the molality, and the mole fraction of ethylene glycol.

Need a reminder? The key ideaShow

Mass percent $= \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$ . Mole fraction $x_A = \dfrac{n_A}{n_A + n_B}$ . Molarity $= \dfrac{\text{mol of solute}}{\text{L of solution}}$ . Molality $= \dfrac{\text{mol of solute}}{\text{kg of solvent}}$ , which does not change with temperature.

The worked solution

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Antifreeze has ethylene glycol, C₂H₆O₂ (62 g mol⁻¹), dissolved in water…

The worked solution

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