CUALES SON LAS FRACCIONES MOLARES DE METANOL Y AGUA EN UNA DISOLUCION QUE CONTIENE 40.0 G DE CH3OH Y 54.0 G DE H2O
Respuestas
Respuesta dada por:
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Mm CH3OH = 32g/mol
H2O = 18 g/mol
n soluto = masa/Mm
n₁ soluto = 40 g / 32 g/mol = 1.25 mol
n₂ solvente = 54 g / 18 g/mol = 3 mol
X soluto = n₁
``````
n₁ + n₂
X soluto = 1.25 mol
```````````````` = 0.30
(1.25 + 3) mol
Xsolvente = n₂
````````
n₁ + n₂
X SOLVENTE = 3 mol
`````````````` = 0.70
(1.25 + 3 ) mol
X SOLUTO + X SOLVENTE = 1
0.30 + 0.70 = 1
H2O = 18 g/mol
n soluto = masa/Mm
n₁ soluto = 40 g / 32 g/mol = 1.25 mol
n₂ solvente = 54 g / 18 g/mol = 3 mol
X soluto = n₁
``````
n₁ + n₂
X soluto = 1.25 mol
```````````````` = 0.30
(1.25 + 3) mol
Xsolvente = n₂
````````
n₁ + n₂
X SOLVENTE = 3 mol
`````````````` = 0.70
(1.25 + 3 ) mol
X SOLUTO + X SOLVENTE = 1
0.30 + 0.70 = 1
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