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27 December, 14:04

A mixture contains 25 g of cyclohexane (C6H12) and 44 g of 2-methylpentane (C6H14). The mixture of liquids is at 35 oC. At this temperature, the vapor pressure of pure cyclohexane is 150 torr, and that of pure 2-methylpentane is 313 torr. Assume this is an ideal solution. What is the mole fraction of cyclohexane in the liquid phase?

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  1. 27 December, 14:20
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    The mol fraction of cyclohexane in the liquid phase is 0.368

    Explanation:

    Step 1: Data given

    Mass of cyclohexane = 25.0 grams

    Mass of 2-methylpentane = 44.0 grams

    Temperature = 35.0 °C

    The pressure of cyclohexane = 150 torr

    The pressure of 2-methylpentane = 313 torr

    The pressure we only need for the mole fraction in gas phase.

    Step 2: Calculate moles of cyclohexane

    Moles cyclohexane = mass cyclohexane / molar mass

    Moles cyclohexane = 25.0 g / 84 g/mol = 0.298 mol of cyclohexane

    Step 3: Calculate moles of 2-methylpentane

    Moles = 44.0 grams / 86 g/mol = 0.512 mol of 2-methylpentane

    Step 4: Calculate mole fraction of cyclohexane in the liquid phase

    Mole fraction of C6H12:

    0.298 / (0.298 + 0.512) = 0.368

    The mol fraction of cyclohexane in the liquid phase is 0.368
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