Question

The vapor pressure of ethanol, CH3 CH2 OH, at 40.0 °C is 17.88 kPa. If 2.24 g of ethanol is enclosed in a 3.00 L container, how much liquid will be present?​

Answer 1

To determine the amount of liquid present, we need to use the ideal gas law equation and the vapor pressure of ethanol at 40.0 °C. By converting the given mass of ethanol into moles and using the ideal gas law equation, we can calculate the volume of the gas. Subtracting this gas volume from the total volume of the container gives us the amount of liquid present.

Step-by-step explanation:

To determine the amount of liquid that will be present, we need to use the ideal gas law equation and the vapor pressure of ethanol at 40.0 °C. We can start by converting the given mass of ethanol (2.24 g) into moles using the molar mass of ethanol (46.07 g/mol). This gives us a value of 0.0486 mol.

Next, we can use the ideal gas law equation, PV = nRT, where P is the vapor pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the equation to solve for V, we get V = nRT/P.

Substituting the values, we have V = (0.0486 mol)(0.0821 L·atm/mol·K)(40.0 °C + 273.15 K)/(17.88 kPa), which gives us a value of approximately 0.257 L. Therefore, the amount of liquid present in the 3.00 L container will be 3.00 L - 0.257 L = 2.743 L.

Answer 2

Based on the mass of ethanol provided, the mass of liquid ethanol is 1.61 g

What is the relationship between gas volume, pressure, temperature and moles?

The relationship between the gas volume, pressure, temperature and moles is given by the ideal gas equation:

• PV = nRT

where:

• P is pressure
• V is volume
• n is number of moles
• R is molar gas constant = 8.314 L.KPa/Kmol
• T is temperature

From data provided:

P = 17.88 kPa

V = 3.00 L

T = 40.0 °C = 313 K

R = 0.049 mole

n = ?

n = PV/RT

n = 17.88 * 2/8.314 * 313

n = 0.0137 moles

Mass of ethanol gas = number of moles * molar mass

Mass of gaseous ethanol = 0.0137 mole * 46 g/mol

Mass of gaseous ethanol = 0.63 g

Then:

Mass of liquid ethanol, CH₃CH₂OH = 2.24 - 0.63

Mass of liquid ethanol, CH₃CH₂OH = 1.61 g

Therefore, the mass of liquid ethanol is 1.61 g

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