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17 April, 16:43

In a 66.0-g aqueous solution of methanol, CH 4 O, CH4O, the mole fraction of methanol is 0.300. 0.300. What is the mass of each component?

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  1. 17 April, 18:27
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    The solution is composed by 37.5 g of water and 28.5 g of methanol.

    Explanation:

    Total mass = 66 g

    Mole fraction methanol: 0.3

    Sum of mole fraction = 1

    Therefore, mole fraction of water = 0.7

    Let's find out the mass of each component by this two equations:

    Methanol mass + Water mass = 66 g

    Water mass = 66g - Methanol mass

    Methanol mass = 66g - water mass

    water mass / 18 g/mol = moles of water

    methanol mass / 32 g/mol = moles of methanol

    moles of water / total moles = 0.7

    moles of methanol / total moles = 0.3

    water mass / 18 g/mol / (water mass / 18 g/mol) + (methanol mass / 32 g/mol) = 0.7; let's replace methanol mass, as (66 - water mass)

    water mass / 18 g/mol / (water mass / 18 g/mol) + (66 - water mass / 32 g/mol) = 0.7 → The unknown is water mass (X)

    X / 18 / ((X / 18) + ((66-X) / 32)) = 0.7

    (X / 18) + ((66-X) / 32) = (16X + 594 - 9X) / 288

    X / 18 / (16X + 594 - 9X) / 288 = 0.7

    X/18 = 0.7. ((7X + 594) / 288)

    X / 18 = 7/2880 (7X + 594)

    X = 7/2880 (7X + 594) 18

    X = 7/160 (7X + 594)

    X = 49/160X + 2079/80

    X - 49/160X = 2079/80

    111/160X = 2079/80

    X = 2079/80. 160/111 = 37.5 g → mass of water

    Therefore, mass of methanol → 66 g - 37.4 g = 28.5 g
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