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?

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

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

where:

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

Learn more about vapor pressure and mass at:

7.13 L of ethanol, CH₃CH₂OH

Explanation:

We'll begin by calculating the number of mole in 2.24 g of ethanol. This can be obtained as follow:

Mass of CH₃CH₂OH = 2.24 g

Molar mass of CH₃CH₂OH = 12 + (3×1) + 12 + (2×1) + 16 + 1

= 12 + 3 + 12 + 2 + 16 + 1

= 46 g/mol

Mole of CH₃CH₂OH =?

Mole = mass / molar mass

Mole of CH₃CH₂OH = 2.24 / 46

Mole of CH₃CH₂OH = 0.049 mole

Next, we shall convert 40 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 40 °C

T(K) = 40 °C + 273

T(K) = 313 K

Finally, we shall determine the volume of ethanol, CH₃CH₂OH in present in the container. This can be obtained as follow:

Pressure (P) = 17.88 KPa

Temperature (T) = 313 K

Number of mole (n) = 0.049 mole

Gas constant (R) = 8.314 L.KPa/Kmol

Volume (V) =?

PV = nRT

17.88 × V = 0.049 × 8.314 × 313

Divide both side by 17.88

V = (0.049 × 8.314 × 313) / 17.88

V = 7.13 L

Thus, 7.13 L of ethanol, CH₃CH₂OH will be present in the container.

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