Answer to Question #24739 in Physical Chemistry for Paul

Question

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.

Answer 1

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