(a) Calculate the density of dinitrogen tetroxide gas $\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\( at \)111.5 \mathrm{kPa}\( and \)0{ }^{\circ} \mathrm{C}$. (b) Calculate the molar mass of a gas if 2.70 g occupies \(0.97 \mathrm{~L}\) at \(134.7 \mathrm{~Pa}\) and \(100^{\circ} \mathrm{C}\).

First, let's convert the given temperature and pressure to Kelvin and Pascals, respectively: Temperature = \(0^{\circ}C + 273.15 = 273.15 K\) Pressure = \(111.5 kPa * 1000 = 111500 Pa\) Next, we need to find the molar mass of dinitrogen tetroxide (N2O4). Using the periodic table, we can find the atomic masses of nitrogen (N) and oxygen (O): Molar mass of nitrogen (N) = 14.01 g/mol Molar mass of oxygen (O) = 16.00 g/mol So, the molar mass of N2O4 will be: Molar mass of N2O4 = 2(14.01) + 4(16.00) = 28.02 + 64.00 = 92.02 g/mol Now, let's rearrange the ideal gas law formula to solve for the density (\(ρ\)) where density = mass / volume: \(ρ = \frac{n \times M}{V}\) \(ρ = \frac{P \times M}{R \times T}\) (density of the gas) Plug in the values: \(ρ = \frac{111500 \times 92.02}{8.314 \times 273.15}\) Calculate the density: \(ρ = \frac{10254711}{2250.88} = 4.56 g/L\) The density of N2O4 at 111.5 kPa and 0°C is \(4.56 g/L\).

First, let's convert the given temperature and pressure to Kelvin and Pascals, respectively: Temperature = \(100^{\circ} C + 273.15 = 373.15 K\) Pressure = \(134.7 Pa\) (already given in Pascals) Now, let's use the ideal gas law to calculate the number of moles (n): \(PV = nRT\) Solve for n: \(n = \frac{PV}{RT}\) Plug in the values: \(n = \frac{134.7 \times 0.97}{8.314 \times 373.15}\) Calculate the number of moles: \(n = \frac{130.459}{3102.47} = 0.04204 \, mol\) Now, let's use the number of moles (n) to calculate the molar mass (M) of the unknown gas. We know that the mass of the gas is 2.70 g. So: Molar mass (M) = mass / moles Solve for M: \(M = \frac{2.70}{0.04204}\) Calculate the molar mass: \(M = 64.23 g/mol\) The molar mass of the unknown gas is \(64.23 g/mol\).