S.28 1-8

Ratkaise atomien luku määrä, kun
a) n( $Cu$ )=0,50 mol

$n=\frac{N}{N_A}$

$0{,}50mol=\ \frac{N}{6{,}022\cdot10^{23}\frac{g}{mol}}$
$N=0{,}50mol\cdot6.022\cdot10^{23}\frac{g}{mol}=3{,}011\cdot10^{23}$
b) n( $C$ )=40 mmol
$40mmol=0{,}04mol$

$n=\frac{N}{N_A}$

$0{,}04mol=\frac{N}{6{,}022\cdot10^{23}\frac{g}{mol}}$
$N=0{,}04mol\cdot6{,}022\cdot10^{23}\frac{g}{mol}=2{,}4088\cdot10^{22}\$
c) n( $O_2$ )= 4,0 mol

$n=\frac{N}{N_A}$

$4{,}0mol=\frac{N}{6{,}022\cdot10^{23}\frac{g}{mol}}\cdot2$
$N=4{,}0mol\cdot6{,}022\cdot10^{23}\frac{g}{mol}\cdot2=4{,}8176\cdot10^{24}$