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The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of an ideal gas. The formula is expressed as \(PV = nRT\). Here's what each symbol stands for:

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant, typically valued at \(0.0821\, L \cdot atm/(mol \cdot K)\).
• \(T\) is the temperature in Kelvin.

To find the molecular mass of a gas, we often rearrange this equation. The rearranged formula \(M = \frac{dRT}{P}\) accounts for the molar mass \(M\), density \(d\), and the previously mentioned constants \(R\), \(T\), and \(P\). When performing such calculations, ensure each substitute value matches the required unit so the result is accurate.

The density of vapor refers to the mass per unit volume of a gaseous substance. In our problem, it is provided as \(1.585 \ g/L\). Density plays an essential role in calculating the molar mass of a vaporized compound. To use it in the Ideal Gas Law, you can substitute the density into the modified formula, \(M = \frac{dRT}{P}\), to calculate the molecular weight of a vapor. It's critical to ensure that the density value is consistent with units used in the Ideal Gas Law, which is usually grams per liter (\(g/L\)). Always consider that density can vary with temperature and pressure, making it crucial to maintain clear control over these conditions during the calculation.

Temperature conversion to Kelvin is a straightforward process crucial for gas law calculations. The Kelvin scale is an absolute temperature scale, which means it starts at absolute zero. To convert a Celsius temperature to Kelvin, use the formula: \[T(K) = T(^\circ C) + 273.15\] In the exercise provided, the temperature is \(90^{\circ}C\), so converting it to Kelvin involves: \[T = 90 + 273.15 = 363.15\, K\] Using Kelvin ensures the proportions in the Ideal Gas Law match, as this scale is directly related to the energy content of the gas. Always double-check conversions to avoid errors in further calculations.

Pressure conversion is another crucial step when working with the Ideal Gas Law. Pressure can be measured in several units, including millimeters of mercury (mmHg) and atmospheres (atm). In many scientific computations, pressure is required in atmospheres, hence the need for conversion. To convert from mmHg to atm, use this simple equation: \[P(\text{atm}) = \frac{P(\text{mmHg})}{760}\] For instance, the exercise gives pressure as \(753\, mmHg\). Converting: \[P = \frac{753}{760} = 0.9908\, atm\] This conversion is essential as it ensures consistency with the ideal gas constant \(R\), which is typically given with atm as the unit of pressure. Proper conversion leads to accurate scientific calculations.