calcular a) la molaridad y b) la molalidad de una disolucion de acido sulfurico de peso especifico 1.198 que contiene 27% de h2so4 en peso
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Mm H2SO4 = 98 g/mol
1 litro = 270 ml de H2SO4
d = m x v
masa = 1.198 g/mL x 270 mL = 323.46 g
M = 1.198 g x 1000 mL x 1 mol x 27 g
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1 mL 1 L 98 g 100 g
M = 3.30 mol/L
mol = 323.46 g / 98 g/mol = 3.30 mol ácido
masa de solvente = 1198 g - 323.46 g = 875.54 g / 1000 = 0.87554 Kg
m = 3.30 mol / 0.87554 Kg
m = 3.769 mol/Kg svte
1 litro = 270 ml de H2SO4
d = m x v
masa = 1.198 g/mL x 270 mL = 323.46 g
M = 1.198 g x 1000 mL x 1 mol x 27 g
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1 mL 1 L 98 g 100 g
M = 3.30 mol/L
mol = 323.46 g / 98 g/mol = 3.30 mol ácido
masa de solvente = 1198 g - 323.46 g = 875.54 g / 1000 = 0.87554 Kg
m = 3.30 mol / 0.87554 Kg
m = 3.769 mol/Kg svte
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