Calculate to three significant digits the density of boron trifluoride gas at exactly 25°C and exactly 1atm. You can assume boron trifluoride gas behaves as an ideal gas under these conditions.

To calculate the density of boron trifluoride gas at exactly 25°C and exactly 1atm, we can use the ideal gas law equation, which is PV = nRT.

Step 1: Convert the temperature to Kelvin. The Kelvin scale starts at absolute zero, which is -273.15°C. So, 25°C is 25 + 273.15 = 298.15 K.

Step 2: Calculate the molar mass of boron trifluoride (BF3). The molar mass of boron (B) is about 10.81 g/mol, and the molar mass of fluorine (F) is about 18.998 g/mol. Since there are three fluorine atoms in boron trifluoride, the molar mass of boron trifluoride is 10.81 g/mol + 3 * 18.998 g/mol = 67.79 g/mol.

Step 3: Substitute the values into the ideal gas law equation. We want to find the density, which is mass/volume. We can express mass as n (number of moles) times the molar mass (M), and volume as PV/RT. So, the density (d) is nM/ (PV/RT) = nRTM/PV = PM/RT.

Step 4: Substitute the values into the density equation. The pressure (P) is 1 atm, the molar mass (M) is 67.79 g/mol, the gas constant (R) is 0.0821 Latm/(Kmol), and the temperature (T) is 298.15 K. So, the density is (1 atm * 67.79 g/mol) / (0.0821 Latm/(Kmol) * 298.15 K) = 2.748 g/L.

Step 5: Round to three significant digits. The density of boron trifluoride gas at exactly 25°C and exactly 1atm is 2.75 g/L.