How many grams of KI are needed to prepare 2,000. grams of an aqueous solution containing 25 parts per million (ppm) of solute?

Question

How many grams of KI are needed to prepare 2,000. grams of an aqueous solution
containing 25 parts per million (ppm) of solute?

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Thanh Thu 2 weeks 2021-09-05T14:58:45+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-09-05T15:00:32+00:00

    Answer:

    m_{solute}=0.05g

    Explanation:

    Hello there!

    In this case, according to the equation for the calculation of the parts per million:

    ppm=\frac{m_{solute}}{m_{solution}} *1x10^6

    It is possible to solve for the mass of KI (solute) as shown below:

    m_{solute}=\frac{ppm*m_{solution}}{1x10^6}

    Thus, we plug in to obtain:

    m_{solute}=\frac{25*2,000g}{1x10^6} \\\\m_{solute}=0.05g

    Best regards!

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