un acido nitrico concentrado de densidad 1 405 g/ml,contiene 68.1% en peso de hno3 calcular a.molaridad
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Mm HNO3 = 63 g /mol
M = 1.405 g x 1000 cm³ x 68.1 g x 1 mol HNO3
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1 cm³ 1 Litro 100 g 63 g HNO3
M = 15.187 molar
M = 1.405 g x 1000 cm³ x 68.1 g x 1 mol HNO3
```````````` `````````````` `````````` ``````````````````
1 cm³ 1 Litro 100 g 63 g HNO3
M = 15.187 molar
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