1- 9 Por.definición.se.sabe.que :
ρ =
mt
=
m H 2O + ms1 VH 2O + V s1
Vt
Como.dato.se.sabe.que :
ρ r = 1,93, ......................luego.la.densidad .del .sólido.es : =
ρ r
ρ s1 ρ H O
→ ρ s1 =
= 1, 93 x1
ρr x ρ H O 2
2
ρ r = 1, 93
g mL
g
mL Por.otro.lado :
ρ s1
=
ms1 Vs1
→ Vs1 =
ms1
23 g
=
ρ s1
1,93 g / mL
= 11,92mL
Vs1 = 11, 92mL Luego : V1 = V H 2O + Vs1 → VH O = V1 − Vs1 = (75 − 11, 92) mL = 63, 08mL 2 V H2O = 63,08mL Cálculo.dela . .masa.de.agua : Suponiendo : ρ H 2O = 1
ρ H O
=
m H 2O
→ mH
V H 2O
2
g mL 2O
=
ρ H O xVH O 2
2
=1
g mL
x 63, 08mL = 63, 08 g
m H2O = 63,08 g Por.tanto : mt1 = mH2O + ms1 = (63, 08 + 23, 0) g = 86, 08 g mt 1 = 86,08 g Luego :
ρ 1 =
1kg 1mL (100cm)3 kg = = 1,14773 x x x = 1147,73 Vt 75,0mL mL 1000 g 1cm3 1m3 m3
mt
86, 08 g
ρ 1 = 1147,73 ρ1 = ρ 2
=
ρ3
ρ = 1147,73
g
kg m
=
3
ρ = 1147,73
kg m3
Cálculo de la fracción 3:
kg m
3
1- 11
1 1
23 75
1
2 2 2
3 3 3
35, 6 83, 2
29,38 80
0,30667
0, 42788
0,36727
1
1
1000
1
(100 3
1
1
1000
1
1
1
)
3
(100 3
306, 67
3
1
) 3
3
3
427,88
3
