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v etanol = 350 mL
m ácido benzoico: 56 g
calcular moles de a. benzoico = n = m/Mm
n = 56 g / 122 g/mol = 0.459 mol
d = m/v
m soluto = 0.789 g/mL x 350 mL = 276.15 g
m = 276.15 /1000 = 0.27615 Kg
m = moles soluto / kg de solvente
m = 0.459 mol / 0.27615 Kg
m = 1.66