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