Odpowiedzi

2011-07-25T14:44:42+02:00

M(CaCl₂)=40g/mol+2*35,5g/mol=111g/mol

m=33,3g

Vr=300cm³=0,3dm³

 

n=m/M

n=33,3g / 111g/mol

n=0,3mol

 

Cm=n/Vr

Cm=0,3mol/0,3dm³

Cm=1mol/dm³

2011-07-25T14:45:59+02:00

CaCl₂

MCaCl₂=40g+2·35,5g=110g

 

110g - 1 mol

33,3g - x mol

____________________

x=n=0,303mol

 

V=300cm²=0,3dm³

 

Cmol=\frac{n}{V}=\frac{0,303mol}{0,3dm^{3}}=1,01\frac{mol}{dm^{3}}\approx1\frac{mol}{dm^{3}}