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A water treatment tablet contains 20.0 mg of tetraglycine hydroperiodide, 40.0% of which is available as soluble iodine. If two tablets are used to treat 1.00 liter of drinking water, what is the concentration in ppm of iodine in the treated water

Sagot :

boffeemadrid

Answer:

[tex]16\ \text{ppm}[/tex]

Explanation:

Mass of one tablet = 20 mg

Mass of two tablets = [tex]2\times 20=40\ \text{mg}[/tex]

Percent that is soluble in water = 40%

Mass of tablet that is soluble in water = [tex]0.4\times 40=16\ \text{mg}[/tex]

So, mass of solute is [tex]16\ \text{mg}[/tex]

Density of water = 1 kg/L

Volume of water = 1 L

So, mass of 1 L of water is [tex]1\times 1=1\ \text{kg}=1000\ \text{g}[/tex]

PPM is given by

[tex]\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\times 10^6=\dfrac{16\times 10^{-3}}{1000}\times 10^6\\ =16\ \text{ppm}[/tex]

Hence, the concentration of iodine in the treated water [tex]16\ \text{ppm}[/tex].

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