Question

If $10.00\mathrm{g}$ of water is introduced into an evacuated flask of volume $2.500\mathrm{L}$ at ${65}^{\circ }\mathrm{C},$ calculate the mass of water vaporized. (Hint: Assume that the volume of the remaining liquid water is negligible; the vapor pressure of water at ${65}^{\circ }\mathrm{C}$ is $187.5\mathrm{mmHg}.$ )

Short answer

0.4015 g of water vaporized at 65°C.

Step-by-step solution

Convert Vapor Pressure to Atmospheres

First, we need to convert the given vapor pressure of water from mmHg to atmospheres because standard calculations use atmospheres. The conversion is done by dividing the pressure value by 760 mmHg/atm. $$P_{atm}=\frac{187.5}{760}=0.2467\text{atm}$$

Use the Ideal Gas Law to Find Moles of Vapor

Next, use the Ideal Gas Law formula, $PV=nRT$, where $P$ is the pressure in atm, $V$ is the volume in liters (2.500 L), $R$ is the gas constant $0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$, and $T$ is the temperature in Kelvin. Convert the temperature to Kelvin first by adding 273 to the Celsius temperature.$$T=65+273=338\text{K}$$Plug these values into the equation to find the number of moles ($n$):$$0.2467\times 2.500=n\times 0.0821\times 338$$Solve for $n$:$$n=\frac{0.2467\times 2.500}{0.0821\times 338}=0.0223\text{moles}$$

Calculate the Mass of Vaporized Water

Now convert the moles of water vapor to grams using the molar mass of water (18.015 g/mol). Multiply the moles by the molar mass:$$m_{vaporized}=0.0223\text{moles}\times 18.015\text{g/mol}=0.4015\text{g}$$

Evaluate Remaining Mass

Finally, evaluate the remaining mass of the water that did not vaporize by subtracting the mass that vaporized from the initial mass:$$m_{remaining}=10.00\text{g}-0.4015\text{g}=9.5985\text{g}$$

Key concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a particular temperature. As the temperature increases, the kinetic energy of liquid molecules increases, allowing more molecules to escape into the vapor phase. This results in a higher vapor pressure. At the given temperature of 65°C, the vapor pressure of water is 187.5 mmHg. Understanding vapor pressure helps in predicting how much liquid might turn into a vapor at a certain temperature. For our calculations, this vapor pressure needs to be expressed in a standard unit, such as atm, for ease of integration into the Ideal Gas Law. This conversion lets us calculate other important properties, like the amount of the substance in moles.

Conversion of Units

One of the fundamental skills in chemistry is converting units to ensure consistency in calculations. For example, vapor pressure given in mmHg needs to be converted to atmospheres (atm) for use in gas law equations. To do this, we use the conversion factor 1 atm = 760 mmHg. In our problem, the vapor pressure of water is 187.5 mmHg. Dividing by 760, we find that this equals 0.2467 atm.

Conversions are not just limited to pressure. Temperature needs to be in Kelvin for calculations involving the Ideal Gas Law. Here, the given temperature of 65°C is converted to Kelvin by adding 273 to get 338 K. These conversions are crucial for maintaining uniformity in calculations and are applicable across all fields of science and engineering.

Moles Calculation

The Ideal Gas Law, expressed as $PV=nRT$, is a cornerstone of physical chemistry. Using this equation, we can find the number of moles () present in a given volume at a specific pressure and temperature. In our example, we know:

• Pressure, $P=0.2467$ atm,
• Volume, $V=2.500$ L,
• Gas constant, $R=0.0821$ L atm K${}^{-1}$ mol${}^{-1}$,
• Temperature, $T=338$ K.

By rearranging the ideal gas equation, we solve for $n$ (moles):$$n=\frac{PV}{RT}$$This yields approximately 0.0223 moles of water vaporized. Understanding how to manipulate this equation allows you to predict the behavior of gases quantitatively under various conditions.

Molar Mass

Molar mass is a measure of the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For water, the molar mass is 18.015 g/mol, derived from adding the atomic masses of two hydrogen atoms (2 x 1.008 g/mol) and one oxygen atom (16.00 g/mol). Knowing the molar mass allows us to easily convert between grams of a substance and moles. In this scenario, we calculated that the moles of vaporized water is 0.0223. By multiplying the number of moles by the molar mass of water, we find that the mass of water vaporized is 0.4015 g. This conversion is frequently used in chemical equations and reactions to transition between mass and moles, ensuring accurate stoichiometric calculations.