Question #24739

The stock solution for this experiment contains 0.7218g of KNO3 in 1000cm3 of water. the manual claims that this solution represents a 100ppm solution of NO3-N. how do I prove this.

Expert's answer

100 ppm is 100 parts per million.

w of NO_{3}^{-} = (N of NO_{3}^{-}) / ( N of H_{2}O) , N is number ofparticles.

0.7218g N of NO_{3}^{-}

KNO_{3} -------> NO_{3}^{-} +

K^{+}

101.1 6.02*10^23

N of NO_{3}^{-} = 0.7218g * 6.02*10^23 / 101.1 = 0.04298 * 10^23

N of H_{2}O = 1000 g *6.02*10^23 / 18 = 334.44 *10^23

w = 0.04298 * 10^23 /334.44 *10^23 = 1.29*10^-4 parts per one part of H_{2}O

1,29*10^-4 * 1000000 = 129 ppm

w of NO

0.7218g N of NO

KNO

K

101.1 6.02*10^23

N of NO

N of H

w = 0.04298 * 10^23 /334.44 *10^23 = 1.29*10^-4 parts per one part of H

1,29*10^-4 * 1000000 = 129 ppm

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