What is the molecular weight of \( 28.5 \mathrm{~g} \) of a gas which occupies \( 2.82 \mathrm{~L} \) at \( 24.9 \mathrm{~atm} \) and \( 436^{\circ} \mathrm{C} \) ?

Step 1 ：Step 1: Calculate the number of moles using Ideal Gas Law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant and \( T \) is temperature in Kelvin. Convert \( 436^{\circ} \mathrm{C} \) to Kelvin by adding 273.15: \(436 + 273.15 = 709.15 \mathrm{K}\)

Step 2 ：Step 2: Solve for \(n\): \( n = \frac{PV}{RT} \). Use \( P = 24.9 \mathrm{~atm} \), \( V = 2.82 \mathrm{~L} \), \( R = 0.0821 \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \), and \( T = 709.15 \mathrm{K} \), then calculate \(n\)

Step 3 ：Step 3: Calculate the molecular weight, \(MW\), using the mass and the number of moles: \( MW = \frac{mass}{n} \). Use \(mass = 28.5 \mathrm{~g}\) and the calculated value of \(n\), then find the \(MW\)