The vapor pressure of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) at \(19^{\circ} \mathrm{C}\) is 40.0 torr. \(\mathrm{A} 1.00\) -g sample of ethanol is placed in a 2.00 \(\mathrm{L}\) container at \(19^{\circ} \mathrm{C}\) . If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol remain?

Recall that the Ideal Gas Law is given by the equation \[PV = nRT\] where P is the pressure, V is the volume, n is the number of moles of the gas, R is the gas constant, and T is the temperature. We are given the value of P as the vapor pressure of ethanol (40.0 torr) at a temperature of \(19^{\circ} \mathrm{C}\). First, we should convert the pressure into atmospheres to be consistent with the gas constant. \(1 \: \text{atm} = 760 \: \text{torr}\), so, \[P = \frac{40.0 \: \text{torr}}{760 \: \text{torr/atm}} = 0.0526 \: \text{atm}\] Next, we need to convert the temperature to Kelvin: \[T = 19^{\circ} \mathrm{C} + 273.15 = 292.15 \: \text{K}\] Now we have everything we need to solve for the moles of ethanol in the vapor phase, using the Ideal Gas Law. Rearrange the equation to solve for n: \[n = \frac{PV}{RT}\]

Now we can plug the given values and constants into the rearranged Ideal Gas Law equation: \[n = \frac{(0.0526 \: \text{atm}) (2.00 \: \text{L})}{(0.0821 \: \text{L atm/mol K})(292.15 \: \text{K})} = 0.00348 \: \text{mol}\]

Now that we have calculated the moles of ethanol in the vapor phase (0.00348 mol), we can convert it to mass using the molar mass of ethanol. The molar mass of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), is approximately 46.07 g/mol. Using the conversion factor: \[0.00348 \: \text{mol} \times \frac{46.07 \: \text{g}}{1 \: \text{mol}} = 0.160 \: \text{g}\]

We initially had 1.00 g of ethanol in the container. Now that we know the mass of ethanol in the vapor phase (0.160 g), we can subtract this value from the initial mass to find the mass of liquid ethanol remaining: \[1.00 \: \text{g} - 0.160 \: \text{g} = 0.840 \: \text{g}\] Therefore, 0.840 g of liquid ethanol remains in the container when the system reaches equilibrium.