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19 December, 15:56

What is the composition of a methanol (CH3OH) - propanol (C3H7OH) solution that has a vapor pressure of 174 torr at 40ºC? At 40ºC, the vapor pressures of pure methanol and pure propanol are 303 and 44.6 torr, respectively. Assume the solution is ideal. Report the composition of the liquid solution in units of mole fractions.

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  1. 19 December, 15:58
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    Psol = Pmeth * Xmeth + Pprop * Xprop

    Where,

    Psol = vapor pressure of the solution

    Pmeth = vapor pressure of methanol

    Pprop = vapor pressure of propanol

    Xmeth = mole fraction methanol

    Xprop = mole fraction propanol

    Therefore,

    174 = 303 Xmeth + 44.6 Xprop

    Xmeth + Xprop = 1

    Xmeth = 1 - Xprop

    174 = 303 (1 - Xprop) + 44.6 Xprop

    174 = 303 - 303 Xprop + 44.6 Xprop

    129 = 258.4 * Xprop

    Xprop = 0.499

    Xmeth = 1 - 0.499

    = 0.501

    Since Xprop = 0.499,

    But Xprop = moles propanol / (moles propanol + moles methanol)

    Since, moles propanol + moles methanol = 1,

    moles of propanol, mprop = 0.499

    moles methanol, mmeth = 0.501

    Molar mass of propanol = (12*3) + (1*8) + 16

    = 60 g/mol.

    mass of propanol, Mprop = 0.499 x 60 g/mol

    = 3.00 g

    Molar mass of methanol = (12*1) + (1*4) + 16

    = 32 g/mol.

    mass of methanol, Mmeth = 0.501 x 32

    = 16.1 g.
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