(a) Calculate the density of sulfur hexafluoride gas at 94.26 \(\mathrm{kPa}\) and \(21{ }^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at $12{ }^{\circ} \mathrm{C}\( and \)99.06 \mathrm{kPa}$..

To calculate the density of sulfur hexafluoride, we'll use the Ideal Gas Law formula: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, convert the given pressure and temperature to the appropriate SI units: Pressure (P) = 94.26 kPa Temperature (T) = 21 °C + 273.15 = 294.15 K Additionally, we'll need to know the molar mass of sulfur hexafluoride (SF6). From the periodic table, the molar mass of sulfur is 32 g/mol, and the molar mass of fluorine is 19 g/mol. Therefore, the molar mass of sulfur hexafluoride is: Molar mass of SF6 = (\(1\times32\)) + (\(6\times19\)) = 32 + 114 = 146 g/mol Now we can use the Ideal Gas Law formula in terms of density (\(\rho\)): \(\rho = \frac{P\times M}{R\times T}\) Where \(\rho\) is the density, P is the pressure, M is the molar mass, R is the ideal gas constant, and T is the temperature in Kelvin. The value of R is 8.314 J/(mol K), but for this calculation, we'll use R = 0.0821 L atm / (mol K) (to match the pressure unit).

Now we can fill in the values: \(\rho = \frac{(94.26\,\text{kPa}) \frac{101.3\,\text{Pa}}{1\,\text{kPa}} \times \frac{1\, \mathrm{atm}}{101325\,\text{Pa}} \times (146\,\text{g/mol})}{(0.0821\,\text{L atm} /(\text{mol K})) \times (294.15\, \text{K})}\) Calculate the density: \(\rho \approx 6.115\, \text{g/L}\) The density of sulfur hexafluoride gas at 94.26 kPa and 21°C is approximately 6.115 g/L.

As given, we'll need to calculate the molar mass of a vapor with: Density (\(\rho\)) = 7.135 g/L Temperature (T) = 12 °C + 273.15 = 285.15 K Pressure (P) = 99.06 kPa We can rearrange the Ideal Gas Law formula in terms of molar mass (M): \(M = \frac{\rho \times R \times T}{P}\)

Now we can fill in the values: \(M = \frac{(7.135\, \text{g/L}) \times (0.0821\, \text{L atm} /(\text{mol K})) \times (285.15\, \text{K})}{(99.06\,\text{kPa}) \frac{101.3\,\text{Pa}}{1\,\text{kPa}} \times \frac{1\, \mathrm{atm}}{101325\,\text{Pa}}}\) Calculate the molar mass: \(M \approx 153.1\, \text{g/mol}\) The molar mass of the vapor that has a density of 7.135 g/L at 12°C and 99.06 kPa is approximately 153.1 g/mol.