Aufgabe 5

clear all;
Mx = 23 %g/mol
Mx = 23
m = 69 %g
m = 69
N_2 = 4.173e24
N_2 = 4.1730e+24
N_A = 6.022e23
N_A = 6.0220e+23

Stoffmenge in Mol

n = m / Mx
n = 3

Teilchenzahl

N = n*N_A
N = 1.8066e+24

Stoffmenge in Mol aus Teilchenzahl

n_2 = N_2 / N_A
n_2 = 6.9296

Deren Masse

m_2 = Mx * n_2
m_2 = 159.3806