The legal limit for drunkenness is 0.08 ( 0.08 grams of alcohol per 100ml of blood). expired air contains the following main gaseous components by volume: 74.1% N2; 15.0% O2; 6.0% H2O; 0.9% Ar; and 4.0% CO2 at 37 degrees Celcius and 1 atm. Determine the density of ethanol.

To determine the density of ethanol, we need to know the molar mass of ethanol. The molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol.

First, we need to calculate the mole fraction of ethanol in the expired air. The mole fraction, denoted by the symbol "X," is the amount of moles of a component divided by the total number of moles of all components.

Given the mole fraction of CO2 in expired air is 4.0%, we can calculate it as follows:

Mole fraction of CO2 = 0.04

Now, let's find the mass of CO2 in 100 ml of expired air. Assuming the density of expired air is equal to 1.0 g/ml, we can calculate the mass of CO2 as follows:

Mass of CO2 = Volume of expired air * Mole fraction of CO2 * Molar mass of CO2
= 100 ml * 0.04 * 44.01 g/mol (molar mass of CO2)
= 176.04 g

Next, we need to determine the amount of ethanol in the expired air corresponding to a blood alcohol level of 0.08 grams per 100 ml of blood. The alcohol level in breath is considered to be equivalent to the alcohol level in the blood.

Since the density of blood is approximately 1.05 g/ml, we can calculate the mass of ethanol in 100 ml of blood as follows:

Mass of ethanol in 100 ml of blood = blood alcohol level per 100 ml of blood / density of blood
= 0.08 / 1.05
≈ 0.0762 g

Now, let's calculate the mole fraction of ethanol in the expired air using the calculated mass of ethanol:

Mole fraction of ethanol = Mass of ethanol in 100 ml of blood / Molar mass of ethanol
= 0.0762 g / 46.07 g/mol
≈ 0.001654

To find the density of ethanol, we can use the ideal gas law, which states that the density of a gas is given by the formula:

Density of gas = (Molar mass of gas * Pressure) / (Gas constant * Temperature)

Assuming the pressure is 1 atm and the temperature is 37 degrees Celsius (which is 310 Kelvin), we can calculate the density of ethanol as follows:

Density of ethanol = (46.07 g/mol * 1 atm) / (0.0821 L·atm/mol·K * 310 K)
≈ 1.071 g/L

Therefore, the density of ethanol is approximately 1.071 g/L.

To determine the density of ethanol, you need to know its molar mass. The molar mass of ethanol (C2H5OH) can be calculated by adding up the molar masses of its constituent elements, carbon (C), hydrogen (H), and oxygen (O).

The molar mass of carbon (C) is 12.01 g/mol.
The molar mass of hydrogen (H) is 1.01 g/mol.
The molar mass of oxygen (O) is 16.00 g/mol.

So, the molar mass of ethanol (C2H5OH) can be calculated as:
(2 * molar mass of carbon) + (6 * molar mass of hydrogen) + (1 * molar mass of oxygen)

= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 24.02 g/mol + 6.06 g/mol + 16.00 g/mol
= 46.08 g/mol

Now that we have the molar mass of ethanol, we can calculate its density using the Ideal Gas Law. The Ideal Gas Law equation is:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant (0.0821 L * atm / K * mol)
T = Temperature (in Kelvin)

In this case, we are given the composition of the expired air, which includes 4.0% CO2. We will assume that the alcohol vapor is ideally mixed with the other gases and that the pressure (P), volume (V), and temperature (T) are constant. We can then write the equation as:

(0.04 x 1 atm) V = n x (0.0821 L * atm / K * mol) x (37 + 273) K

Simplifying the equation, we have:

0.04 V = n x 0.0821 x 310.15
0.04 V = 2.5411 n

Now, let's calculate the number of moles of ethanol (n) present:

n = (0.08 g / 46.08 g/mol) = 0.001737 mol

Substituting the value of n into the equation, we have:

0.04 V = 2.5411 x 0.001737
0.04 V = 0.0044
V = 0.0044 / 0.04
V = 0.11 L

Now that we have the volume, we can calculate the density (d) using the formula:

d = mass / volume

Since we already know the molar mass of ethanol is 46.08 g/mol, we can calculate the mass as:

mass = n x molar mass
mass = 0.001737 mol x 46.08 g/mol
mass = 0.08 g

Now substituting the calculated values into the density formula, we get:

d = 0.08 g / 0.11 L = 0.73 g/L

Therefore, the density of ethanol is 0.73 g/L.