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Exercise · Q26

Q.A solution is prepared by mixing 36 g36\ \text{g} of water (H2O\text{H}_2\text{O}, molar mass 18.02 g mol−118.02\ \text{g mol}^{-1}) with 46 g46\ \text{g} of ethanol (C2H5OH\text{C}_2\text{H}_5\text{OH}, molar mass 46.07 g mol−146.07\ \text{g mol}^{-1}). Calculate the mole fraction of water and of ethanol in the solution.

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Moles of water =3618.02=1.998≈2.00 mol= \dfrac{36}{18.02} = 1.998 \approx 2.00\ \text{mol}

Moles of ethanol =4646.07=0.9985≈1.00 mol= \dfrac{46}{46.07} = 0.9985 \approx 1.00\ \text{mol}

Total moles =2.00+1.00=3.00 mol= 2.00 + 1.00 = 3.00\ \text{mol}

xwater=2.003.00=0.667,xethanol=1.003.00=0.333x_{\text{water}} = \frac{2.00}{3.00} = 0.667, \qquad x_{\text{ethanol}} = \frac{1.00}{3.00} = 0.333 …

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