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16 December, 03:41

What mass of na2co3 is present in 0.650 l of a 0.505 m na2co3 solution?

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  1. 16 December, 06:27
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    V=0.650 L

    c=0.505 mol/L

    M (Na₂CO₃) = 106.0 g/mol

    the amount of substance of sodium carbonate is:

    n (Na₂CO₃) = vc

    the mass of sodium carbonate is:

    m (Na₂CO₃) = n (Na₂CO₃) M (Na₂CO₃)

    m (Na₂CO₃) = vcM (Na₂CO₃)

    m (Na₂CO₃) = 0.650*0.505*106.0=34.7945 g
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