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28 November, 14:58

What mass of NaNO3 is needed to make 3.5 L of 2.2 M solution? (Na = 23.0g, N=14.0g, O=16.0g)

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  1. 28 November, 17:56
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    = 654.5 g NaNO3

    Explanation:

    Molarity = Moles / volume

    Volume = 3.5 L

    Molarity = 2.2 M

    Therefore;

    Moles = Molarity * volume

    = 3.5 L * 2.2 M

    = 7.7 moles

    But 1 mole of NaNO3 is 85 g/mol

    Mass of NaNO3 = 85 g/mole * 7.7 moles

    = 654.5 g NaNO3
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