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Molar mass is a fundamental concept in chemistry, referring to the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. It is expressed in units of grams per mole (g/mol). This concept is crucial when we need to convert between the mass of a substance and the number of moles, as the two are related by the molar mass.

For example, in the calculation given in the exercise, the molar mass of argon is 39.95 g/mol. This means that 39.95 grams of argon is equivalent to one mole of argon atoms. To find the number of moles ( ext{n}) in a particular mass of argon, we use the formula:
\[ n = \frac{m}{M} \]
where \( m \) is the mass in grams and \( M \) is the molar mass.
Given 175 grams of argon, the calculation becomes:
\[ n = \frac{175\, g}{39.95\, g/mol} \approx 4.38\, ext{moles} \]
This conversion step is essential for any further calculations involving the ideal gas law.

Pressure calculation is an important application of the ideal gas law, which is an equation of state for a hypothetical gas known as an "ideal gas." The law describes how the pressure of a gas relates to its volume, temperature, and number of moles.

The equation is given by:
\[ PV = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant (0.08206 \( L \cdot atm/mol \cdot K \)), and \( T \) is the temperature in Kelvin.

To find the pressure when the temperature is 225 K, we rearrange the ideal gas law to solve for \( P \):
\[ P = \frac{nRT}{V} \]
Substitute in the known values:
- \( n = 4.38 \) moles
- \( R = 0.08206 \text{ L}\cdot \text{atm/mol}\cdot \text{K} \)
- \( T = 225 \) K
- \( V = 2.50 \) L

This leads to:
\[ P = \frac{4.38 \times 0.08206 \times 225}{2.50} \approx 16.14\, ext{atm} \]
Thus, when the temperature is 225 K, the pressure in the container is approximately 16.14 atm.

The temperature calculation using the ideal gas law involves determining the temperature of a gas when its pressure, volume, and number of moles are known. In the exercise, we use this to find the temperature when the pressure is 10.0 atm.

We again refer to the ideal gas law which states:
\[ PV = nRT \]
To solve for temperature \( T \), we rearrange the equation to get:
\[ T = \frac{PV}{nR} \]
Given the values:
- \( P = 10.0 \) atm
- \( V = 2.50 \) L
- \( n = 4.38 \) moles
- \( R = 0.08206 \text{ L}\cdot \text{atm/mol}\cdot \text{K} \)

Substitute these into the equation:
\[ T = \frac{10.0 \times 2.50}{4.38 \times 0.08206} \approx 694\, ext{K} \]
Therefore, the temperature when the pressure is 10.0 atm is approximately 694 Kelvin, illustrating how changes in pressure can impact gas temperature when using the ideal gas law.