Question

Blood alcohol content is usually measured as the mass of ethanol (C2H5OH, FM = 46.068 g/mole) per volume of blood. More specifically, the results are reported in g/dL. If the density of blood is 1.06 g/mL what is the molality and molarity of a BAC that is 0.15 g/dL?

Answer 1

D = 1.06 g/ml

find molality of a BAC 0.15 g/dL

molality = mol of alcohgol / kg solvent

assume a basis of 100 g or 0.1 Kg

V = m/D = 100/1.06 = 94.33962 ml = 0.9433962 dL

then

Find mass of ethanol

mass = C*V = 0.15 g / dL * 0.9433962 = 0.1415 g

mol = mass/MW = 0.1415/46.068 = 0.00307 mol of ethanol

then

molality = 0.00307/0.1 =  0.0307 molal

b)

molarity

M = mol/L

mol =  0.00307 mol of ethanol

L = 0.9433962 dL = 0.09433962 L

then

M = 0.00307 /0.09433962 = 0.032542 mol of ethanol per liter