Question

Calculate to three significant digits the density of boron trifluoride gas at exactly -10 °C and exactly 1 atm. You can assume boron trifluonde gas behaves as an ideal gas under these conditions.

Answer 1

The density of boron trifluoride gas at -10 °C and 1 atm is calculated using the ideal gas law to be 3.14 g/L to three significant digits.

Step-by-step explanation:

To calculate the density of boron trifluoride gas at -10 °C and 1 atm, we first need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature. Therefore, -10 °C is equivalent to 263.15 K. We can use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin.

Since we need to find the density (ρ), which is mass per unit volume, we need the molar mass of boron trifluoride (BF3). The atomic masses of boron (B) and fluorine (F) are 10.81 amu and 19.00 amu respectively. The molar mass of BF3 is thus (1 × 10.81) + (3 × 19.00) = 67.81 g/mol.

With that, the ideal gas law can be rearranged to solve for molar volume (V/n), which is V/n = RT/P. Substituting in the values for R, T, and P gives us the molar volume. The density of the gas can then be calculated using ρ = molar mass / molar volume.

Using these values:

1. Calculate molar volume: V/n = (0.0821 L·atm/K·mol) × (263.15 K) / (1 atm) = 21.6 L/mol (rounded to three significant digits).
2. Calculate density: ρ = (67.81 g/mol) / (21.6 L/mol) = 3.14 g/L (to three significant digits).

Therefore, the density of boron trifluoride gas at -10 °C and 1 atm is 3.14 g/L.