Final answer:
Using the ideal gas law, the density of 1.655 g/dm³ at 323K and 1.01×10⁵ Pa allows calculation of the molar mass Mr of the gas by rearranging the formula to M = ρRT/P and substituting the values with the appropriate unit conversions.
Step-by-step explanation:
To calculate the molar mass (Mr) of a gas given its density, temperature, and pressure, we can use the ideal gas law, which is shown in the equation PV = nRT. Here, the density (ρ) can be related to the molar mass (M) and molar volume (Vm) by ρ = M/Vm, and Vm can be expressed in terms of the ideal gas law as Vm = RT/P. By substituting these into each other, we can solve for Mr.
Given the density of 1.655 g/dm³ (which is the same as g/L), temperature (T) of 323K, and pressure (P) of 1.01×10⁵ Pa, we want to find Mr of the gas. We can rearrange the molar mass formula to M = ρRT/P. Using the gas constant R = 8.31 J/mol·K (as it is convenient for the SI units given), we plug in the values to get M = (1.655 g/L)(8.31 J/mol·K)(323K)/(1.01×10⁵ Pa).
However, to ensure the units are consistent for M (which should be in g/mol), remember that 1 J = 1 kg·m²/s², so we need to convert the pressure from Pa to kg/m·s² and the gas constant R from J/mol·K to kg·m²/s²·mol·K. After adjusting the units and calculating, we arrive at the molar mass Mr of the gas.