Calculate to three significant digits the density of boron trifluoride gas at exactly \( 20^{\circ} \mathrm{C} \) and exactly 1 atm. You can assume boron trifluoride gas behaves as an ideal gas under these conditions. \( 1.57 \frac{\mathrm{~g}}{\mathrm{~L}} \)

To calculate the density of boron trifluoride (BF₃) gas at exactly \( 20^{\circ} \mathrm{C} \) and \( 1 \, \text{atm} \), we'll use the ideal gas law. Here's a step-by-step breakdown of the calculation: ### **1. Ideal Gas Law Overview** The ideal gas law is given by: \[ PV = nRT \] Where: - \( P \) = Pressure (1 atm) - \( V \) = Volume (in liters) - \( n \) = Number of moles - \( R \) = Ideal gas constant (\( 0.0821 \, \text{L·atm/mol·K} \)) - \( T \) = Temperature in Kelvin ### **2. Rearranging for Density** Density (\( \rho \)) is mass per unit volume (\( \rho = \frac{m}{V} \)). To relate this to the ideal gas law: \[ n = \frac{m}{M} \] Where \( M \) is the molar mass of BF₃. Substituting \( n \) into the ideal gas equation: \[ PV = \frac{m}{M}RT \implies \rho = \frac{m}{V} = \frac{PM}{RT} \] So, \[ \rho = \frac{PM}{RT} \] ### **3. Calculating Molar Mass of BF₃** - **Boron (B):** \( 10.81 \, \text{g/mol} \) - **Fluorine (F):** \( 18.998 \, \text{g/mol} \) (since there are 3 F atoms in BF₃) \[ M = 10.81 + 3 \times 18.998 = 10.81 + 56.994 = 67.804 \, \text{g/mol} \] ### **4. Converting Temperature to Kelvin** \[ T = 20^{\circ} \mathrm{C} + 273.15 = 293.15 \, \text{K} \] ### **5. Plugging Values into the Density Formula** \[ \rho = \frac{1 \, \text{atm} \times 67.804 \, \text{g/mol}}{0.0821 \, \text{L·atm/mol·K} \times 293.15 \, \text{K}} \] \[ \rho = \frac{67.804}{24.055} \approx 2.82 \, \text{g/L} \] ### **6. Conclusion** The calculated density of boron trifluoride gas under the given conditions is: \[ \boxed{2.82 \, \text{g/L}} \] This value is **approximately twice** the initially proposed value of \( 1.57 \, \text{g/L} \). Given the assumptions of ideal gas behavior and standard calculations, \( 2.82 \, \text{g/L} \) is the accurate density for BF₃ at \( 20^{\circ} \mathrm{C} \) and \( 1 \, \text{atm} \).