Blood alcohol concentration (BAC) is a measure of alcohol in your body, expressed as grams of alcohol per 100ml of blood. For a BAC of 0.05, every 100ml of your blood contains 0.05 grams of alcohol. Your BAC can be measured through your breath, blood, and urine. Drivers with an open licence have to keep under a BAC of 0.05.

Scientists estimate the mass of blood in a human body to be approximately 8 percent of body weight. If we assume that 17 % of the alcohol that a person drinks from beer goes into the bloodstream, how many litres of beer can an 71 kg person drink to get a BAC of 0.05? The alcohol content of beer is 6 % volume, and the blood density is 1025kg/m3, the alcohol density is 789 kg/m3.

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maacastrobr maacastrobr · Answer 1 · 2019-09-22T12:05:18+02:00

Answer:

A 71 kg person will get a BAC of 0.05 when drinking 0.3442 L of beer

Explanation:

The mass of blood in a human body is approximately 8%, so if a person weighs 71 kg, the mass of blood would be:

71 kg * 8/100 = 5.68 kg of blood.

Using blood density, we can calculate the volume that 5.68 kg of blood occupies:

5.68 kg * [tex]\frac{1m^{3}}{1025kg}[/tex] = 0.0055415 m³

We convert m³ into mL, keeping the unit that we want to convert to in the numerator; and the unit that we want to convert in the denominator:

[tex]0.0055415m^{3}*\frac{1000L}{1m^{3}} *\frac{1000mL}{1L}=5541.5mL[/tex]

Now we calculate the amount of alcohol that would be needed in the bloodstream to get a BAC of 0.05:

[tex]\frac{5541.5mL}{100mL}*0.05g[/tex] = 2.77 g of alcohol are needed in the bloodstream in order to have a BAC of 0.05

The amount of alcohol that needs to be ingested is higher than 2.77 g, due to the fact that only 17% of the alcohol goes into the bloodstream, so:

2.77 g [tex]*\frac{100}{17} =[/tex] 16.29 g of alcohol need to be ingested

Then we use the alcohol concentration of beer to calculate the volume of beer needed, using the alcohol density. First we convert the alcohol density to g/L, making sure the units that we want to convert cancel each other:

[tex]789\frac{kg}{m^{3}}*\frac{1000g}{1kg} *\frac{1m^{3}}{1000L} =789g/L[/tex]

Now we use the density to calculate the litres of alcohol needed, keeping in mind that 16.29 g of alcohol are needed:

[tex]16.29g*\frac{1L}{789g}=[/tex] 0.02065 L of alcohol are needed.

Finally we calculate the litres of beer needed, keeping in mind the concentration of alcohol in beer:

[tex]0.02065L_{alcohol}*\frac{100L_{beer}}{6L_{alcohol}} =[/tex]0.3442 L of beer are needed.