Zad. 8
mNaOH=40g/mol
n=40g/40g/mol=1mol
V=230cm3=0,23dm3
Cm=1mol/0,23dm3=4,35mol/dm3
Zad.9
Cm=n/V -> n=V*Cm
n=2mol/dm3*2,5dm3=5moli KNO3
mKNO3=101 g/mol
1mol - 101g
5 moli - x
x=505g KNO3
Zad.10.
120g - 100%
x g - 5%
x=6g
Po dolaniu 300cm3 wody:
mwody=d*V=300g
mr = 120+300g=420g
ms=6g
420g - 100%
6g-x%
x= 1,43%
