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14 January, 10:33

How many grams of solute are present in 625 mL of 0.520 M KBr?

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  1. 14 January, 12:20
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    M (KBr) = 119.0 g/mol

    v=625 mL=0.625 L

    c=0.520 mol/L

    n (KBr) = cv

    m (KBr) = n (KBr) M (KBr)

    m (KBr) = cvM (KBr)

    m (KBr) = 0.520*0.625*119.0=38.675 g

    38.675 g
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