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28 May, 19:51

A sodium nitrate solution is 21.5% (by mass) of nano3 (molar mass = 85.00 g/mol) and the solution has a density of 1.08 g/ml. calculate the molarity (m) of the solution.

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  1. 28 May, 20:18
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    Answer is: molarity of sodium nitrate solution is 2.73 M.

    V (NaNO₃) = 1 L · 1000 mL/L = 1000 mL.

    d (NaNO₃) = 1.08 g/mL.

    ω (NaNO₃) = 21.5% : 100% = 0.215.

    mr (NaNO₃) = V (NaNO₃) · d (NaNO₃).

    mr (NaNO₃) = 1000 mL · 1.08 g/mL.

    mr (NaNO₃) = 1080 g.

    m (NaNO₃) = ω (NaNO₃) · mr (NaNO₃).

    m (NaNO₃) = 0.215 · 1080 g.

    m (NaNO₃) = 232.2 g.

    n (NaNO₃) = m (NaNO₃) : M (NaNO₃).

    n (NaNO₃) = 232.2 g : 85.00 g/mol.

    n (NaNO₃) = 2.73 mol.

    c (NaNO₃) = n (NaNO₃) : V (NaNO₃).

    c (NaNO₃) = 2.73 mol : 1 L.

    c (NaNO₃) = 2.73 mol/L.
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