CALCULE LA FRACCION MOLAR DE LOS COMPONENTES DE UNA MEZCLA DE 20 GRAMOS DE METANOL (CH3OH) EN 90 GRAMOS DE AGUA
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CALCULE LA FRACCION MOLAR DE LOS COMPONENTES DE UNA MEZCLA DE 20 GRAMOS DE METANOL (CH3OH) EN 90 GRAMOS DE AGUA
Mm H2O = 18 g/mol
CH3OH = 32 g/mol
1. calcular moles
mol soluto = 20 g / 32 g/mol = 0.625 mol
mol solvente = 90 g/ 18 g/mol = 5 mol
2. calcular: n total = nsto + nsvte
ntotal = 0.625 mol + 5 mol = 5.625 mol
3. calcvular fraccion molar
Xsto = n sto
````````
n total
Xsto = 0.625 mol
``````````````
5.625 mol
Xsto = 0.111
X svte = n svte
``````````
n total
X svte = 5 mol
``````````
5.625 mol
X svte = 0.888
xSTO + x SVTE = 1
0.111 + 0.888 = 0.999 ≈ 1
Mm H2O = 18 g/mol
CH3OH = 32 g/mol
1. calcular moles
mol soluto = 20 g / 32 g/mol = 0.625 mol
mol solvente = 90 g/ 18 g/mol = 5 mol
2. calcular: n total = nsto + nsvte
ntotal = 0.625 mol + 5 mol = 5.625 mol
3. calcvular fraccion molar
Xsto = n sto
````````
n total
Xsto = 0.625 mol
``````````````
5.625 mol
Xsto = 0.111
X svte = n svte
``````````
n total
X svte = 5 mol
``````````
5.625 mol
X svte = 0.888
xSTO + x SVTE = 1
0.111 + 0.888 = 0.999 ≈ 1
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