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Answer to Question #24739 in Physical Chemistry for Paul

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 NO3- = (N of NO3-) / ( N of H2O) , N is number ofparticles.

0.7218g N of NO3-
KNO3 -------> NO3- +
K+
101.1 6.02*10^23


N of NO3- = 0.7218g * 6.02*10^23 / 101.1 = 0.04298 * 10^23

N of H2O = 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 H2O

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

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