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5 December, 09:40

Prepare a 1.00 L of a 0.25 M solution of sodium carbonate. Determine the amount of sodium carbonate needed in correct significant figures.

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  1. 5 December, 10:33
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    Mass = 26.50 g

    Explanation:

    Given dа ta:

    Mass of sodium carbonate = ?

    Molarity of solution = 0.25 M

    Volume of solution = 1.00 L

    Solution:

    Molarity = number of moles / volume in litter

    0.25 M = number of moles / 1.00 L

    Number of moles = 0.25 M * 1.00 L

    Number of moles = 0.25 mol

    Mass:

    Mass = number of moles * molar mass

    Mass = 0.25 mol * 106 g/mol

    Mass = 26.50 g
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