calcule los gramos de agua que se deben agregar a 8.50 g de H3PO4 para preparar una solución a 0.1 molal
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Mm H3PO4 = 98 g/mol
m = mol soluto
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Kg svte
mol = masa /Mm
mol = 8.5 g / 98 g/mol
mol = 0.086
Kg svte = mol soluto
``````````````
m
Kg svte = 0.086 mol
```````````````
0.1 mol /Kg svte
Kg = 0.86 de agua
g = 0.86 x 1000 = 860 g de agua
m = mol soluto
````````````````
Kg svte
mol = masa /Mm
mol = 8.5 g / 98 g/mol
mol = 0.086
Kg svte = mol soluto
``````````````
m
Kg svte = 0.086 mol
```````````````
0.1 mol /Kg svte
Kg = 0.86 de agua
g = 0.86 x 1000 = 860 g de agua
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