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8 April, 07:27

What mass of potassium bromide (in grams) do you need to make 250.0 mL of a 1.50 M potassium bromide solution?

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  1. 8 April, 07:46
    0
    44.63g

    Explanation:

    First, let us calculate the number of mole of KBr in 1.50M KBr solution.

    This is illustrated below:

    Data obtained from the question include:

    Volume of solution = 250mL = 250/1000 = 0.25L

    Molarity of solution = 1.50M

    Mole of solute (KBr) =.?

    Molarity is simply mole of solute per unit litre of solution

    Molarity = mole / Volume

    Mole = Molarity x Volume

    Mole of solute (KBr) = 1.50 x 0.25

    Mole of solute (KBr) = 0.375 mole

    Now, we calculate the mass of KBr needed to make the solution as follow:

    Molar Mass of KBr = 39 + 80 = 119g/mol

    Mole of KBr = 0.375 mole

    Mass of KBr = ?

    Mass = number of mole x molar Mass

    Mass of KBr = 0.375 x 119

    Mass of KBr = 44.63g

    Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution
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