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18 July, 07:04

Calculate the molality of a 27.0% (by mass) aqueous solution of nitric acid.

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  1. 18 July, 08:43
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    Answer is: molality is 5,86 m.

    ω (HNO₃) = 27,0% = 0,27.

    If we use mass of solution 100 grams:

    m (solution) = 100 g.

    m (HNO₃) = ω (HNO₃) · m (solution).

    m (HNO₃) = 0,27 · 100 g.

    m (HNO₃) = 27 g.

    n (HNO₃) = m (HNO₃) : M (HNO₃).

    n (HNO₃) = 27 g : 63 g/mol.

    n (HNO₃) = 0,428 mol.

    m (H₂O) = 100 g - 27 g.

    m (H₂O) = 73 g = 0,073 kg.

    b (solution) = n (HNO₃) : m (H₂O).

    b (solution) = 0,428 mol : 0,073 kg.

    b (solution) = 5,86 mol/kg = 5,86 m.
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