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29 January, 04:08

A laboratory procedure calls for making 410.0mL of a 1.2M NaNO3 solution.

What mass of NaNO3 (in g) is needed?

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  1. 29 January, 05:09
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    Molar mass NaNO3 = 85 g/mol

    Volume in liters: 410.0 mL / 1000 = > 0.41 L

    mass = M x molar mass x V

    mass = 1.2 x 85 x 0.41

    mass = 41.82 g of NaNO3
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