A 15.0g sample of liquid ethanol, C2H5OH, absorbs 5.13x10^3 J of heat at its normal boiling point, 78.0C. The molar enthalpy of vaporization of ethanol is 39.3kJ/mol. (a) What volume of ethanol vapor is produced? The voume is measured at 78.0C and 1.00atm pressure. (b) What mass of ethanol remians in the liquid state?

Divide 5.13x10^3 J by 39.3kJ/mol. to get the number of moles of ethanol evaporated.

(a) Use the ideal gas low to calculate the volume of that vapor
V = n RT/P

(b) Convert the number of moles vaporized to mass and subtract that from 15.0 g.

To solve this problem, we need to use the equations relating heat, moles, molar mass, and volume.

First, let's determine the number of moles of ethanol in the 15.0g sample.

Step 1: Calculate the number of moles of ethanol.
Molar mass of ethanol (C2H5OH) = (2 x atomic mass of carbon) + (6 x atomic mass of hydrogen) + atomic mass of oxygen
Molar mass of C2H5OH = (2 x 12.01 g/mol) + (6 x 1.01 g/mol) + 16.00 g/mol
Molar mass of C2H5OH = 46.07 g/mol

Number of moles of ethanol = mass of ethanol / molar mass of ethanol
Number of moles of ethanol = 15.0 g / 46.07 g/mol
Number of moles of ethanol = 0.3255 mol

Step 2: Calculate the volume of ethanol vapor using the ideal gas law.

The ideal gas law equation is PV = nRT, where:
P = pressure (1.00 atm)
V = volume (unknown)
n = number of moles of ethanol (0.3255 mol)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (78.0°C + 273.15 = 351.15 K)

Rearranging the ideal gas law equation to solve for V, we get:

V = nRT / P

V = (0.3255 mol) x (0.0821 L·atm/(mol·K)) x (351.15 K) / (1.00 atm)

V = 9.76 L

Therefore, the volume of the ethanol vapor produced is 9.76 L (rounded to two decimal places).

Part (a) is solved.

To determine the mass of ethanol remaining in the liquid state, we need to calculate the amount of heat required to vaporize the 0.3255 mol of ethanol.

Step 1: Calculate the heat required using the molar enthalpy of vaporization.

Heat required (Q) = number of moles of ethanol x molar enthalpy of vaporization
Heat required (Q) = 0.3255 mol x 39.3 kJ/mol

Step 2: Convert the heat required to joules.

1 kJ = 1000 J

Heat required (Q) = (0.3255 mol x 39.3 kJ/mol) x 1000 J/kJ

Heat required (Q) = 12,811.05 J

Therefore, the heat required to vaporize the ethanol is 12,811.05 J.

Step 3: Calculate the remaining mass of ethanol.

Remaining mass of ethanol = initial mass - mass evaporated
Remaining mass of ethanol = 15.0 g - (12,811.05 J / 5.13 x 10^3 J/g)

Remaining mass of ethanol = 15.0 g - 2.50 g

Remaining mass of ethanol = 12.50 g

Therefore, the mass of ethanol remaining in the liquid state is 12.50 g. (rounded to two decimal places).

Part (b) is solved.

To solve this problem, we need to use the concept of molar enthalpy of vaporization and the ideal gas law. Let's break it down into two parts:

(a) To find the volume of ethanol vapor produced, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to calculate the number of moles of ethanol vapor. We know the molar enthalpy of vaporization is 39.3 kJ/mol, so we can convert the heat absorbed to moles by dividing it by the molar enthalpy of vaporization:

5.13 x 10^3 J = 1 mol
? = 39.3 kJ/mol

5.13 x 10^3 J * (1 mol / 39.3 kJ) ≈ 0.131 mol

Now that we have the number of moles (n), we can rearrange the ideal gas law equation to solve for V:

V = nRT / P

First, we need to convert the temperature to Kelvin by adding 273.15:

78.0°C + 273.15 = 351.15 K

Now we can substitute the values into the equation:

V = (0.131 mol)(0.0821 L·atm / K·mol)(351.15 K) / 1.00 atm

V ≈ 3.76 L

Therefore, approximately 3.76 liters of ethanol vapor are produced at 78.0°C and 1.00 atm pressure.

(b) To find the mass of ethanol remaining in the liquid state, we can subtract the mass of ethanol vapor produced from the initial mass of the liquid ethanol.

The molar mass of ethanol (C2H5OH) is:
2(12.01 g/mol) + 6(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol

Using this molar mass, we can calculate the initial moles of ethanol present in the liquid using the initial mass of 15.0 g:

15.0 g * (1 mol / 46.07 g) ≈ 0.325 mol

Since one mole of liquid ethanol produces one mole of ethanol vapor when it evaporates, the remaining moles of ethanol in the liquid state will be the initial moles minus the moles of ethanol vapor:

0.325 mol - 0.131 mol ≈ 0.194 mol

Finally, we can convert the remaining moles of ethanol to mass by multiplying it by the molar mass:

0.194 mol * 46.07 g/mol ≈ 8.94 g

Therefore, approximately 8.94 grams of ethanol remain in the liquid state.