Question #24729

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

Expert's answer

100ppm means 100 parts per million.

W of NO3^- = (N of NO3^-) / ( N of H2O) , N is number of particles.

0.7218g N of NO3^-

& KNO3 -------> NO3- & + K+

& 101.1 6.02*10^23 ( Avogadro constant)

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

So answer is 129 ppm, not 100ppm

W of NO3^- = (N of NO3^-) / ( N of H2O) , N is number of particles.

0.7218g N of NO3^-

& KNO3 -------> NO3- & + K+

& 101.1 6.02*10^23 ( Avogadro constant)

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

So answer is 129 ppm, not 100ppm

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