1. Find the molar mass of BF₃:
* Boron (B) has a molar mass of 10.81 g/mol.
* Fluorine (F) has a molar mass of 19.00 g/mol.
* BF₃ has a molar mass of 10.81 + (3 * 19.00) = 67.81 g/mol
2. Calculate the number of moles of BF₃:
* Moles = mass / molar mass
* Moles = 0.155 g / 67.81 g/mol
* Moles ≈ 0.00229 mol
3. Use the Ideal Gas Law at STP:
* STP (Standard Temperature and Pressure) is defined as 0°C (273.15 K) and 1 atm.
* Ideal Gas Law: PV = nRT
* P = Pressure (1 atm)
* V = Volume (what we want to find)
* n = number of moles (0.00229 mol)
* R = Ideal gas constant (0.0821 L·atm/mol·K)
* T = Temperature (273.15 K)
4. Solve for Volume (V):
* V = (nRT) / P
* V = (0.00229 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
* V ≈ 0.0517 L
Therefore, 0.155 g of boron trifluoride occupies approximately 0.0517 liters at STP.
