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16 June, 03:14

If 8.97 grams of (nh4) 2co3 is used to make a 0.250-molar solution, what is the final volume of the solution?

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  1. 16 June, 06:22
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    0.3733 L or 373 mL

    Solution:

    Data Given:

    Mass = 8.97 g

    M. Mass of (NH₄) ₂CO₃ = 96.09 g. mol⁻¹

    Molarity = 0.250 mol. L⁻¹

    Step 1: Calculate Moles of Ammonium Carbonate as,

    Moles = Mass : M. Mass

    Putting values,

    Moles = 8.97 g : 96.09 g. mol⁻¹

    Moles = 0.0933 mol

    Step 2: Calculate Volume of Solution as,

    Molarity = Moles : Volume

    Solving for Volume,

    Volume = Mole : Molarity

    Putting values,

    Volume = 0.0933 mol : 0.250 mol. L⁻¹

    Volume = 0.3733 L

    Or,

    Volume = 373 mL
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