Question

A 355 -mL container holds 0.146 g of Ne and an unknown amount of Ar at \(35^{\circ} \mathrm{C}\) and a total pressure of \(626 \mathrm{mmHg}\). Calculate the number of moles of Ar present.

Short answer

There are 0.00437 moles of Ar present.

Step-by-step solution

- Converting Volume to Liters

Given the volume of the container is 355 mL, convert this to liters by dividing by 1000. Volume in liters: \[ V = \frac{355 \text{ mL}}{1000} = 0.355 \text{ L} \]

- Converting Celsius to Kelvin

Convert the temperature from Celsius to Kelvin by adding 273.15. Temperature in Kelvin: \[ T = 35^{\text{°C}} + 273.15 = 308.15 \text{ K} \]

- Converting Pressure to Atmospheres

Convert the pressure from mmHg to atmospheres using the conversion factor \(1 \text{ atm} = 760 \text{ mmHg}\). Pressure in atmospheres: \[ P = \frac{626 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.8237 \text{ atm} \]

- Calculating Moles of Ne

Calculate the moles of Ne using its molar mass (Neon has a molar mass of approximately 20.18 g/mol). \[ n_{\text{Ne}} = \frac{0.146 \text{ g}}{20.18 \text{ g/mol}} = 0.00723 \text{ mol} \]

- Using the Ideal Gas Law

Use the Ideal Gas Law to calculate the total number of moles of gas in the container. The Ideal Gas Law is given by \( PV = nRT \). Using \( R = 0.0821 \text{ L·atm·K}^{-1} \text{·mol}^{-1} \), we can solve for \( n \). \[ n_{\text{total}} = \frac{(0.8237 \text{ atm})(0.355 \text{ L})}{(0.0821 \text{ L·atm·K}^{-1} \text{·mol}^{-1})(308.15 \text{ K})} = 0.0116 \text{ mol} \]

- Calculating Moles of Ar

Subtract the moles of Ne from the total moles to find the moles of Ar. \[ n_{\text{Ar}} = n_{\text{total}} - n_{\text{Ne}} = 0.0116 \text{ mol} - 0.00723 \text{ mol} = 0.00437 \text{ mol} \]

Key concepts

gas laws

Gas laws are mathematical relationships that describe the behavior of gases. The Ideal Gas Law is one of the most important because it combines several simpler gas laws. The formula is written as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature. This equation helps to predict how gases will behave under different conditions by relating these variables. Understanding this law allows us to solve for any one of these variables if the others are known.
It is crucial to remember:

• Always use Kelvin for temperature.
• Pressure units must be consistent.
• Volume should be in liters.

Let's dive deeper into some of these conversions.

molar mass

Molar mass is the mass of one mole of a substance. This is typically expressed in grams per mole (\text{g/mol}). Every element on the periodic table has a specific molar mass listed, and it tells you how much one mole of that element weighs. For example, Neon (Ne) has a molar mass of approximately \( 20.18 \text{ g/mol} \).
To calculate the number of moles of a substance, use the formula: \[ n = \frac{m}{M} \] where:

• n is the number of moles
• m is the mass in grams
• M is the molar mass in \text{g/mol}

In the exercise, this calculation was used to find the moles of Ne from its given mass.

unit conversions

Unit conversions are critical in ensuring the values used in equations are consistent. Often, you'll have to convert:

• Volume from milliliters (mL) to liters (L) by dividing by 1000
• Pressure from millimeters of mercury (mmHg) to atmospheres (atm)
• Temperature from degrees Celsius (°C) to Kelvin (K)

For instance, converting the given volume of 355 mL to liters results in 0.355 L as 355 divided by 1000 is 0.355.
Similarly, for pressure conversions: 1 atm is equal to 760 mmHg, so to convert 626 mmHg to atmospheres, you can use the formula: \[ P = \frac{626 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.8237 \text{ atm} \] These conversions are necessary for proper use in the Ideal Gas Law.

temperature conversion

Temperature conversion is essential in gas law calculations as the Ideal Gas Law requires temperature in Kelvin. To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. For example, in the exercise, the temperature was given in Celsius as 35°C. The calculation is:
\[ T = 35^{\text{°C}} + 273.15 = 308.15 \text{ K} \]
This ensures the temperature is in the correct unit required for gas law equations. Kelvin is used because it starts at absolute zero, making it a more natural fit for thermodynamic calculations. Always remember to check the unit you're using for temperature and convert if necessary.