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At 300°C, the vapor pressure of Hg is 32.97 torr. What mass of Au would have to be dissolved in 5.00 g of Hg to lower its vapor pressure to 25.00 torr?

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  • Dr W
    Lv 7
    3 weeks ago

    we use "raoult's law" to solve this.. which states

    .. PA = χ A * (PsatA)

    https://en.wikipedia.org/wiki/Raoult%27s_lawwhere

    .. PA = partial pressure of component A over the solution

    .. χ A = mole fraction of component A in the liquid phase

    .. PsatA = saturated vapor pressure of A.. i.e.. the pressure of A

    .. .. . .. .. ..over the liquid if the liquid was pure component A

    .. .. .. .. . . i.e... the pressure of A over the liquid if χ A = 1

    and via Daltons law of partial pressures, we know that

    .. Ptotal = PA + PB + PC + etc

    *************

    so for this case, we can write

    .. Ptotal = PAu + PHg

    .. Ptotal = χ Au * PsatAu + χ Hg * PsatHg

    since we were only given PsatHg.. (that's the 32.97 torr).. and not PsatAu, we going to assume PsatAu = 0.. i.e.. we assume gold is non-volatile. In which case

    .. Ptotal = χ Hg * PsatHg

    solving for χ Hg

    .. χ Hg = Ptotal / PsatHg = 25.00 torr / 32.97 torr = 0.75826 (1 extra sig fig)

    and we know

    .. ... ... .. .. mol Hg.. ... ... . ... ... 5.00g * 1mol/200.59g

    χ Hg = ---- ----- ----- ----- = ----- ----- ----- ---- ------ ----- ----- --- = 0.75826

    .. ... . . mol Hg + mol Au.. 5.00g * 1mol / 200.59g + mol Au

    solving for mole Au with a bit of algebra while dropping units

    .. mol Au = ((5.00/200.59) / 0.75826) - 5.00/200.59 = 0.007947

    and finally

    .. mass Au = 0.007947 mol * (196.97g / mol) =1.56g.. (3 sig figs)

  • 3 weeks ago

    Au dissolves in Hg to form a solution. At 300°C, the vapor pressure of Au is much lower than that of Hg. Hence, it can be assumed that the solute, Au, is non-volatile.

    Vapor pressure of the pure solvent, P°(Hg) = 32.97 torr

    Vapor pressure of the solution, P = 25.00 torr

    P = X(Hg) * P°(Hg)

    Mole fraction of Hg, X(Hg)= P / P°(Hg) = 25.00 / 32.97 = 0.7583

    Molar mass of Hg = 200.6 g/mol

    Moles of Hg = (5.00 g) * (1 mol / 200.6 g) = 0.02493 mol

    When M mol of Au dissolves in 5.00 g of Hg.

    Mole fraction of Hg, X(Hg) = 0.02493 / (M + 0.02493) = 0.7583

    M + 0.02493 = 0.03288

    M + 0.02493 = 0.02493 / 0.7583

    No. of moles of Au, M = 0.00795 mol

    Molar mass of Au = 197.0 g/mol

    Mass of Au = (0.00795 mol) × (197.0 g / 1 mol) = 1.57 g

    ====

    OR:

    m = mass of Au

    Mole fraction of Hg calculated by using moles of components = Mole fraction of calculated by vapor pressure

    (5.00/200.6) / [(m/197.0) + (5.00/200.6)] = 25.00 / 32.97

    0.02493 / (0.005076m + 0.02493) = 0.7583

    0.02493 = 0.003849m + 0.01890

    0.003849m = 0.006030

    Mass of Au, m = 1.57 g

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