Question

A \(1.20-\mathrm{g}\) sample of water is injected into an evacuated \(5.00-\mathrm{L}\) flask at \(65^{\circ} \mathrm{C}\). What percentage of the water will be vapor when the system reaches equi librium? Assume ideal behavior of water vapor an that the volume of liquid water is negligible. The va por pressure of water at \(65^{\circ} \mathrm{C}\) is \(187.5 \mathrm{mmHg}\).

Short answer

The percentage of the water that will be vapor when the system reaches equilibrium is \(44.6%.\)

Step-by-step solution

Convert grams to moles

Using the molar mass of water (\(18.02 g/mol\)), convert the mass of water from grams to moles. This is done by dividing the given mass by the molar mass. Thus, the number of moles of water (n) equals \(1.20 g / 18.02 g/mol = 0.0666 moles\).

Calculate the total pressure in atm

We are given that the vapor pressure of water at \(65^{\circ}C\) is \(187.5 mmHg\). However, in order to use the ideal gas law, this pressure needs to be converted to atmospheres (atm) from mmHg. There are \(760 mmHg\) in \(1 atm\), therefore \(187.5 mmHg / 760 = 0.2467 atm\).

Calculate the moles of vaporized water

Since the vapor and the liquid are in equilibrium, the vapor is at its partial pressure. Using the ideal gas law equation, \(PV=nRT\), where P is pressure, V is volume, n is number of moles, R is the gas constant (\(0.0821 L·atm/K·mol\)), and T is temperature in kelvins. Solving for n (number of moles), we get: \(n = PV/RT = (0.2467 atm * 5.00 L) / (0.0821 * (65+273)) = 0.0297 moles\). This is the number of moles of water that has vaporized.

Determine the percentage of vaporized water

To find the percentage of water that has vaporized, divide the moles of the vapor by the total moles of the sample and multiply by 100. Thus, the percentage is \((0.0297/0.0666)*100 = 44.6%\). So, \(44.6%\) of the water vaporizes.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental concept in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is represented by the formula: \( PV = nRT \), where \( P \) represents the pressure of the gas, \( V \) is the volume it occupies, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) represents the temperature in Kelvin.

For the water vapor in our exercise, we applied the ideal gas law to calculate the number of moles of water that has vaporized at equilibrium. By manipulating the formula to solve for \( n \), as in \( n = \frac{PV}{RT} \), we could deduce the amount of gas present in moles. It's important to note that the ideal gas law assumes that the gas behaves 'ideally,' meaning intermolecular forces are negligible and the gas particles occupy no volume, which is a good approximation for many gases under standard conditions.

Convert Grams to Moles

To understand the chemistry of reactions, it is essential to convert mass (in grams) to moles, as reactions happen on a molecular level, and moles reflect the number of particles rather than mass. The conversion factor is the molar mass of a substance.

In this exercise, converting grams of water to moles was the first step. As we know, the molar mass of water is approximately \( 18.02 g/mol \). This conversion is done by dividing the given mass by the water's molar mass, symbolized by the equation \( n = \frac{mass}{molar \ mass} \). Hence, for our water sample, \( n = \frac{1.20 g}{18.02 g/mol} \), yielding \( 0.0666 \) moles of water.

Calculate Vaporization Percentage

Determining the vaporization percentage involves comparing the amount of a substance that has changed states (from liquid to gas in this case) to the total amount present initially.

In our exercise, we found that \( 0.0297 \) moles of the water vaporized. To find the vaporization percentage, we divided this by the total number of moles (originally in liquid form) and then multiplied by 100 to obtain a percentage: \( (0.0297/0.0666)*100 = 44.6% \). This calculation tells us that \(44.6%\) of the water in the flask has vaporized when the system reached equilibrium, demonstrating the dynamic nature of the phase change in a closed system.