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11 April, 23:44

Calculate the molality of a 25.4% (by mass) aqueous solution of phosphoric acid (H3PO4).

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  1. 12 April, 01:34
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    W=0.254 (25.4%)

    M (H₃PO₄) = 98.0 g/mol

    m - the mass of the solution

    m (H₃PO₄) = mw

    n (H₃PO₄) = m (H₃PO₄) / M (H₃PO₄)

    m (H₂O) = m (1-w)

    the molality is:

    Cm=n (H₃PO₄) / m (H₂O)

    Cm=mw/[M (H₃PO₄) m (1-w) ]=w/[M (H₃PO₄) (1-w) ]

    Cm=0.254/[98 * (1-0.254) ]=0.003474 mol/g=3.474*10⁻³ mol/g=3.474 mol/kg
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