mass of mixture = 106.26kg/hr
10% of the mixture is ethanol = 10.626 kg
number of moles of ethanol = "\\dfrac{10626\\text{g} }{46\\text{g\/mol}}"
= 231 moles
90% of the mixture is water = 95.634 kg
number of moles of water = "\\dfrac{95634\\text{g}}{18\\text{g\/mol}}"
= 5313 moles
Total number of moles = 231 + 5313 = 5544 moles
mole fraction of ethanol = 231/5544 = 0.0417
mole fraction of water = 5313/5544 = 0.9583
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