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19 April, 21:02

What is the concentration of bromide, in ppm, if 12.41 g MgBr2 is dissolved in 2.55 L water.

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  1. 19 April, 22:07
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    concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

    Explanation:

    ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

    In our case we have:

    mass of MgBr₂ = 12.41 g

    volume of water (which is equal to the final solution volume) = 2.55 L

    Now we devise the following reasoning:

    if 12.41 g of MgBr₂ are dissolved in 2.55 L of water

    then X g of MgBr₂ are dissolved in 1 L of water

    X = (1 * 12.41) / 2.55 = 4.867 g of MgBr₂

    if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻

    then in 4.867 g of MgBr₂ we have Y g of Br⁻

    Y = (4.867 * 160) / 184 = 4.232 g of bromide (Br⁻)

    4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

    concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
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